Some Basic Concepts in Chemistry MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Identifying the unchanged gas component:}\) Nitrogen is explicitly stated to be inert. Therefore, its amount remains \[ n_{\mathrm{N_2,final}}=0.30\,\mathrm{mol} \] The hydrogen formed by the reaction is the only other dry gas. \(\textbf{Using the measured final mole fraction:}\) Let \(x\) be the amount of hydrogen produced. The total dry gas amount is \[ n_{\mathrm{gas,total}}=0.30+x \] The hydrogen mole fraction is \[ X_{\mathrm{H_2}}=\frac{x}{0.30+x} \] Substituting the measured value, \[ 0.40=\frac{x}{0.30+x} \] \[ 0.40(0.30+x)=x \] \[ 0.12+0.40x=x \] \[ 0.60x=0.12 \] \[ x=0.20\,\mathrm{mol} \] Therefore, \[ n_{\mathrm{H_2,formed}}=0.20\,\mathrm{mol} \] \(\textbf{Reconstructing hydrochloric acid consumption:}\) The balanced equation consumes two moles of HCl for every mole of hydrogen formed. \[ n_{\mathrm{HCl,consumed}}=2(0.20) \] \[ n_{\mathrm{HCl,consumed}}=0.40\,\mathrm{mol} \] All supplied HCl is stated to have reacted. Thus, \[ n_{\mathrm{HCl,initial}}=0.40\,\mathrm{mol} \] \(\textbf{Determining zinc consumption:}\) The zinc-to-hydrogen mole ratio is \(1:1\). \[ n_{\mathrm{Zn,consumed}}=0.20\,\mathrm{mol} \] The zinc mass consumed is \[ m_{\mathrm{Zn,consumed}}=0.20(65) \] \[ m_{\mathrm{Zn,consumed}}=13.0\,\text{g} \] \(\textbf{Checking the complete dry-gas composition:}\) The final gas amount is \[ n_{\mathrm{gas,total}}=0.30+0.20=0.50\,\mathrm{mol} \] The hydrogen mole fraction is \[ X_{\mathrm{H_2}}=\frac{0.20}{0.50}=0.40 \] This reproduces the observed composition. The inert nitrogen changes the final gas mole fraction but does not consume HCl or alter the chemical equation. The reaction's stoichiometry links the inferred hydrogen amount to the acid and zinc consumed. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{HCl,initial}}=0.40\,\mathrm{mol},\quad m_{\mathrm{Zn,consumed}}=13.0\,\text{g}} \]

Explanation: \(\textbf{Representing the two carbonate amounts:}\) Let \(x\) moles be sodium carbonate and \(y\) moles be sodium hydrogen carbonate. Their combined formula-unit amount is \[ x+y=0.10 \qquad (1) \] \(\textbf{Using the different acid requirements:}\) Each mole of sodium carbonate consumes two moles of HCl. Each mole of sodium hydrogen carbonate consumes one mole of HCl. Since both react completely and no HCl remains, \[ 2x+y=0.16 \qquad (2) \] \(\textbf{Recovering the original mixture composition:}\) Subtracting Equation (1) from Equation (2), \[ x=0.16-0.10 \] \[ x=0.06\,\mathrm{mol} \] Therefore, \[ y=0.10-0.06 \] \[ y=0.04\,\mathrm{mol} \] \(\textbf{Calculating sodium carbonate mole fraction:}\) \[ X_{\mathrm{Na_2CO_3}}=\frac{0.06}{0.10} \] \[ X_{\mathrm{Na_2CO_3}}=0.60 \] \(\textbf{Tracking sodium chloride production:}\) Each mole of sodium carbonate produces two moles of NaCl. Its contribution is \[ n_{\mathrm{NaCl,from\ carbonate}}=2(0.06) \] \[ n_{\mathrm{NaCl,from\ carbonate}}=0.12\,\mathrm{mol} \] Each mole of sodium hydrogen carbonate produces one mole of NaCl. Its contribution is \[ n_{\mathrm{NaCl,from\ hydrogen\ carbonate}}=0.04\,\mathrm{mol} \] The total is \[ n_{\mathrm{NaCl,total}}=0.12+0.04 \] \[ n_{\mathrm{NaCl,total}}=0.16\,\mathrm{mol} \] \(\textbf{Converting sodium chloride amount to mass:}\) \[ M_{\mathrm{NaCl}}=23+35.5 \] \[ M_{\mathrm{NaCl}}=58.5\,\mathrm{g\,mol^{-1}} \] Therefore, \[ m_{\mathrm{NaCl}}=0.16(58.5) \] \[ m_{\mathrm{NaCl}}=9.36\,\text{g} \] \(\textbf{Independent carbon dioxide check:}\) Both reactions produce one carbon dioxide molecule per reacting carbonate formula unit. The combined amount of carbon dioxide is therefore \[ n_{\mathrm{CO_2}}=0.06+0.04=0.10\,\mathrm{mol} \] The total HCl consumption is \[ 2(0.06)+0.04=0.16\,\mathrm{mol} \] Both specified experimental constraints are satisfied. The two substances produce the same amount of carbon dioxide per mole of formula units but consume different amounts of acid. That difference provides the information needed to determine their individual amounts. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{Na_2CO_3}}=0.60,\quad m_{\mathrm{NaCl}}=9.36\,\text{g}} \]

Explanation: \(\textbf{Establishing the decomposed formula-unit amount:}\) The formula \(\mathrm{NH_4HCO_3}\) contains one nitrogen atom, five hydrogen atoms, one carbon atom and three oxygen atoms. Its molar mass is \[ M_{\mathrm{NH_4HCO_3}}=14+5(1)+12+3(16) \] \[ M_{\mathrm{NH_4HCO_3}}=79\,\mathrm{g\,mol^{-1}} \] Therefore, \[ n_{\mathrm{NH_4HCO_3}}=\frac{7.9}{79} \] \[ n_{\mathrm{NH_4HCO_3}}=0.100\,\mathrm{mol} \] \(\textbf{Recovering the three decomposition products:}\) The balanced equation produces equal mole amounts of ammonia, carbon dioxide and water. Thus, \[ n_{\mathrm{NH_3}}=0.100\,\mathrm{mol} \] \[ n_{\mathrm{CO_2}}=0.100\,\mathrm{mol} \] \[ n_{\mathrm{H_2O}}=0.100\,\mathrm{mol} \] \(\textbf{Applying the first separation:}\) All water vapour is condensed and removed. Its mass is \[ m_{\mathrm{H_2O}}=0.100(18) \] \[ m_{\mathrm{H_2O}}=1.8\,\text{g} \] Therefore, the mass of the dry gas mixture is \[ m_{\mathrm{dry\ gases}}=7.9-1.8 \] \[ m_{\mathrm{dry\ gases}}=6.1\,\text{g} \] \(\textbf{Independent dry-gas mass calculation:}\) The ammonia molar mass is \[ M_{\mathrm{NH_3}}=14+3(1)=17\,\mathrm{g\,mol^{-1}} \] The carbon dioxide molar mass is \[ M_{\mathrm{CO_2}}=44\,\mathrm{g\,mol^{-1}} \] Their combined mass is \[ m_{\mathrm{dry\ gases}}=0.100(17)+0.100(44) \] \[ m_{\mathrm{dry\ gases}}=1.7+4.4 \] \[ m_{\mathrm{dry\ gases}}=6.1\,\text{g} \] \(\textbf{Applying the second separation:}\) The absorber removes all ammonia and none of the carbon dioxide. Therefore, the gas leaving it contains only \[ n_{\mathrm{CO_2,outlet}}=0.100\,\mathrm{mol} \] \(\textbf{Calculating the final STP volume:}\) \[ V_{\mathrm{CO_2,outlet}}=nV_m \] \[ V_{\mathrm{CO_2,outlet}}=0.100(22.4) \] \[ V_{\mathrm{CO_2,outlet}}=2.24\,\text{L} \] \(\textbf{Distinguishing the two separations:}\) Condensation removes water from the original three-component gaseous product mixture. The ammonia absorber then removes a second component from the already dried gases. The initial hot gas amount, dry gas amount and final outlet gas amount are therefore different. The volume of gas leaving the absorber must not include either removed water vapour or absorbed ammonia. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{dry\ gases}}=6.1\,\text{g},\quad V_{\mathrm{CO_2,outlet}}=2.24\,\text{L}} \]

Explanation: \(\textbf{Separating the two necessary consistency tests:}\) A correctly recorded complete decomposition must satisfy overall mass conservation. It must also satisfy the fixed stoichiometric relationship between the pure carbonate and its two pure products. Meeting total mass conservation alone is therefore not sufficient. \(\textbf{Establishing the theoretical product-mass relationship:}\) The molar masses are \[ M_{\mathrm{CaCO_3}}=100\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{CaO}}=56\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{CO_2}}=44\,\mathrm{g\,mol^{-1}} \] Therefore, every 100 g of completely decomposed pure calcium carbonate produces 56 g CaO and 44 g carbon dioxide. The required product-mass ratio is \[ m_{\mathrm{CaO}}:m_{\mathrm{CO_2}}=56:44 \] \[ m_{\mathrm{CaO}}:m_{\mathrm{CO_2}}=14:11 \] \(\textbf{Testing Trial P:}\) For 10.0 g carbonate, the required product masses are \[ m_{\mathrm{CaO}}=10.0\left(\frac{56}{100}\right)=5.6\,\text{g} \] \[ m_{\mathrm{CO_2}}=10.0\left(\frac{44}{100}\right)=4.4\,\text{g} \] Both agree with the record. \(\textbf{Testing Trial Q:}\) For 20.0 g carbonate, \[ m_{\mathrm{CaO}}=11.2\,\text{g} \] \[ m_{\mathrm{CO_2}}=8.8\,\text{g} \] Again, both agree with the record. \(\textbf{Testing Trial R:}\) For 15.0 g carbonate, the expected oxide mass is \[ m_{\mathrm{CaO,expected}}=15.0\left(\frac{56}{100}\right) \] \[ m_{\mathrm{CaO,expected}}=8.4\,\text{g} \] The expected carbon dioxide mass is \[ m_{\mathrm{CO_2,expected}}=15.0\left(\frac{44}{100}\right) \] \[ m_{\mathrm{CO_2,expected}}=6.6\,\text{g} \] The recorded values are 8.0 g and 7.0 g. Their sum is still 15.0 g, so overall mass conservation appears satisfied. However, the individual product amounts contradict the required fixed stoichiometric composition. \(\textbf{Testing Trial S:}\) For 25.0 g carbonate, \[ m_{\mathrm{CaO}}=25.0\left(\frac{56}{100}\right)=14.0\,\text{g} \] \[ m_{\mathrm{CO_2}}=25.0\left(\frac{44}{100}\right)=11.0\,\text{g} \] These match the record. \(\textbf{Scientific conclusion:}\) Only Trial R fails the stoichiometric product-distribution test. A reported experiment can preserve total mass while assigning incorrect masses to individual products. The law of conservation of mass and the fixed composition implied by a balanced reaction provide complementary checks.

Explanation: \(\textbf{Using a common gas-volume basis:}\) At the same temperature and pressure, gas volumes are proportional to molecular amounts. Therefore, the stated fractional conversion of oxygen molecules can be applied directly to its initial reference volume. \(\textbf{Determining oxygen consumed:}\) The initial oxygen volume is 9.0 L. Exactly two-thirds is converted. \[ V_{\mathrm{O_2,consumed}}=9.0\left(\frac23\right) \] \[ V_{\mathrm{O_2,consumed}}=6.0\,\text{L} \] The unreacted oxygen volume is \[ V_{\mathrm{O_2,remaining}}=9.0-6.0 \] \[ V_{\mathrm{O_2,remaining}}=3.0\,\text{L} \] \(\textbf{Recovering the ozone volume:}\) The balanced equation gives \[ 3\mathrm{O_2}\rightarrow2\mathrm{O_3} \] Thus, three gaseous molecular quantities of oxygen produce two quantities of ozone. \[ V_{\mathrm{O_3,formed}}=6.0\left(\frac23\right) \] \[ V_{\mathrm{O_3,formed}}=4.0\,\text{L} \] \(\textbf{Adding the inert gas contribution:}\) Nitrogen does not react. Its amount remains equivalent to a reference volume of 3.0 L. The final gas mixture therefore contains ozone, unreacted oxygen and nitrogen. \[ V_{\mathrm{final}}=4.0+3.0+3.0 \] \[ V_{\mathrm{final}}=10.0\,\text{L} \] \(\textbf{Calculating the ozone mole fraction:}\) At common temperature and pressure, mole fraction equals the corresponding gas-volume fraction. \[ X_{\mathrm{O_3}}=\frac{V_{\mathrm{O_3}}}{V_{\mathrm{final}}} \] \[ X_{\mathrm{O_3}}=\frac{4.0}{10.0} \] \[ X_{\mathrm{O_3}}=0.40 \] \(\textbf{Checking oxygen-atom conservation:}\) The 6.0 reference-volume units of oxygen consumed represent oxygen molecules containing two atoms each. They provide twelve corresponding oxygen-atom units. The 4.0 reference-volume units of ozone contain three oxygen atoms per molecule. They also represent twelve oxygen-atom units. Therefore, the oxygen atoms are conserved even though the number of gaseous molecules decreases during ozone formation. \(\textbf{Final answer:}\) \[ \boxed{V_{\mathrm{final}}=10.0\,\text{L},\quad X_{\mathrm{O_3}}=0.40} \]

Explanation: \(\textbf{Converting oxygen molecules into amount:}\) The supplied oxygen count is \[ N_{\mathrm{O_2}}=3.011\times10^{23} \] The corresponding amount is \[ n_{\mathrm{O_2}}=\frac{N_{\mathrm{O_2}}}{N_A} \] \[ n_{\mathrm{O_2}}=\frac{3.011\times10^{23}}{6.022\times10^{23}} \] \[ n_{\mathrm{O_2}}=0.500\,\mathrm{mol} \] \(\textbf{Using the iron-to-oxygen coefficient ratio:}\) The balanced equation consumes four iron atoms for every three oxygen molecules. Therefore, \[ n_{\mathrm{Fe,consumed}}=0.500\left(\frac43\right) \] \[ n_{\mathrm{Fe,consumed}}=\frac23\,\mathrm{mol} \] \(\textbf{Converting the iron amount into atoms:}\) \[ N_{\mathrm{Fe}}=n_{\mathrm{Fe}}N_A \] \[ N_{\mathrm{Fe}}=\frac23(6.022\times10^{23}) \] \[ N_{\mathrm{Fe}}\approx4.015\times10^{23}\ \text{atoms} \] \(\textbf{Recovering the oxide formula-unit amount:}\) Three moles of oxygen molecules produce two moles of iron(III) oxide formula units. Therefore, \[ n_{\mathrm{Fe_2O_3}}=0.500\left(\frac23\right) \] \[ n_{\mathrm{Fe_2O_3}}=\frac13\,\mathrm{mol} \] \(\textbf{Converting formula-unit amount into count:}\) \[ N_{\mathrm{Fe_2O_3}}=\frac13(6.022\times10^{23}) \] \[ N_{\mathrm{Fe_2O_3}}\approx2.007\times10^{23}\ \text{formula units} \] \(\textbf{Independent constituent-atom check:}\) Every iron(III) oxide formula unit contains two iron atoms. Thus, the number of iron atoms incorporated into the product is \[ N_{\mathrm{Fe\ in\ oxide}}=2N_{\mathrm{Fe_2O_3}} \] \[ N_{\mathrm{Fe\ in\ oxide}}\approx4.015\times10^{23} \] This agrees with the iron atoms consumed. The oxygen molecules initially contain \[ N_{\mathrm{O\ atoms}}=2(3.011\times10^{23}) \] \[ N_{\mathrm{O\ atoms}}=6.022\times10^{23} \] The oxide formula units contain three oxygen atoms each. Their combined oxygen-atom count is \[ 3(2.00733\ldots\times10^{23})=6.022\times10^{23} \] Both elemental inventories are conserved. The calculation distinguishes an iron atom, an oxygen molecule and an ionic oxide formula unit as different counted entities. \(\textbf{Final answer:}\) \[ \boxed{N_{\mathrm{Fe}}\approx4.015\times10^{23},\quad N_{\mathrm{Fe_2O_3}}\approx2.007\times10^{23}} \]

Explanation: \(\textbf{Separating the contributions of the two metals:}\) Both metals produce hydrogen on reaction with hydrochloric acid. However, their metal-to-hydrogen mole ratios differ. Therefore, the measured total hydrogen amount must be combined with the original mixture mass to recover the two component amounts. \(\textbf{Representing the unknown metal amounts:}\) Let \[ n_{\mathrm{Mg}}=x \] \[ n_{\mathrm{Al}}=y \] The initial mixture mass is 5.10 g. Therefore, \[ 24x+27y=5.10 \qquad (1) \] \(\textbf{Using hydrogen stoichiometry:}\) Magnesium produces one mole of hydrogen per mole of metal. Aluminium produces three moles of hydrogen from two moles of metal. Thus, \[ n_{\mathrm{H_2,total}}=x+\frac32y \] The measured hydrogen amount gives \[ x+\frac32y=0.250 \qquad (2) \] \(\textbf{Solving the two independent constraints:}\) From Equation (2), \[ x=0.250-\frac32y \] Substituting into Equation (1), \[ 24\left(0.250-\frac32y\right)+27y=5.10 \] \[ 6.00-36y+27y=5.10 \] \[ 9y=0.90 \] \[ y=0.100\,\mathrm{mol} \] Therefore, \[ x=0.250-\frac32(0.100) \] \[ x=0.100\,\mathrm{mol} \] \(\textbf{Recovering the metal masses:}\) \[ m_{\mathrm{Mg}}=0.100(24) \] \[ m_{\mathrm{Mg}}=2.4\,\text{g} \] \[ m_{\mathrm{Al}}=0.100(27) \] \[ m_{\mathrm{Al}}=2.7\,\text{g} \] Their sum is \[ 2.4+2.7=5.1\,\text{g} \] \(\textbf{Calculating acid consumption:}\) The magnesium reaction consumes two moles of HCl per mole of magnesium. \[ n_{\mathrm{HCl,Mg}}=2(0.100)=0.200\,\mathrm{mol} \] The aluminium reaction consumes three moles of HCl per mole of aluminium. \[ n_{\mathrm{HCl,Al}}=3(0.100)=0.300\,\mathrm{mol} \] Hence, \[ n_{\mathrm{HCl,total}}=0.200+0.300 \] \[ n_{\mathrm{HCl,total}}=0.500\,\mathrm{mol} \] \(\textbf{Independent hydrogen check:}\) Magnesium produces \(0.100\,\mathrm{mol}\) hydrogen. Aluminium produces \(0.150\,\mathrm{mol}\) hydrogen. Their sum is \[ 0.100+0.150=0.250\,\mathrm{mol} \] This reproduces the observed gas amount. The equal mole amounts of the two metals are an inferred result; they must not be assumed from the initial mixture mass. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{Mg}}=2.4\,\text{g},\quad m_{\mathrm{Al}}=2.7\,\text{g},\quad n_{\mathrm{HCl}}=0.50\,\mathrm{mol}} \]

Explanation: \(\textbf{Distinguishing pre-existing product from newly formed product:}\) The vessel initially contains some phosphorus(V) oxide. This product amount must be included in the final inventory, but it does not need to be formed again. The balanced equation determines only the amount newly produced from phosphorus and oxygen. \(\textbf{Testing the oxygen requirement:}\) One mole of \(\mathrm{P_4}\) requires five moles of oxygen. Complete consumption of the initial phosphorus would require \[ n_{\mathrm{O_2,required}}=5(0.20) \] \[ n_{\mathrm{O_2,required}}=1.00\,\mathrm{mol} \] Only \(0.50\,\mathrm{mol}\) oxygen is available. Therefore, oxygen is limiting under the specified reaction conditions. \(\textbf{Determining phosphorus consumption:}\) \[ n_{\mathrm{P_4,consumed}}=\frac{0.50}{5} \] \[ n_{\mathrm{P_4,consumed}}=0.10\,\mathrm{mol} \] The remaining phosphorus amount is \[ n_{\mathrm{P_4,remaining}}=0.20-0.10 \] \[ n_{\mathrm{P_4,remaining}}=0.10\,\mathrm{mol} \] \(\textbf{Calculating the newly produced oxide:}\) The equation forms one mole of phosphorus(V) oxide per mole of phosphorus consumed. \[ n_{\mathrm{P_4O_{10},new}}=0.10\,\mathrm{mol} \] \(\textbf{Combining the final oxide inventory:}\) Initially, the vessel already contained \(0.10\,\mathrm{mol}\) oxide. The final amount is \[ n_{\mathrm{P_4O_{10},final}}=n_{\mathrm{P_4O_{10},initial}}+n_{\mathrm{P_4O_{10},new}} \] \[ n_{\mathrm{P_4O_{10},final}}=0.10+0.10 \] \[ n_{\mathrm{P_4O_{10},final}}=0.20\,\mathrm{mol} \] \(\textbf{Independent phosphorus-atom check:}\) The initial elemental phosphorus contains \[ n_{P,\mathrm{P_4}}=4(0.20)=0.80\,\mathrm{mol} \] The pre-existing oxide contains \[ n_{P,\mathrm{initial\ oxide}}=4(0.10)=0.40\,\mathrm{mol} \] Therefore, the initial total phosphorus-atom amount is \[ 0.80+0.40=1.20\,\mathrm{mol} \] Finally, the remaining elemental phosphorus contains \(4(0.10)=0.40\,\mathrm{mol}\) phosphorus atoms. The final oxide contains \(4(0.20)=0.80\,\mathrm{mol}\) phosphorus atoms. Their sum is again \(1.20\,\mathrm{mol}\). The calculation distinguishes reaction extent from the final amount of a species already present before reaction. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{P_4O_{10},final}}=0.20\,\mathrm{mol},\quad n_{\mathrm{P_4,remaining}}=0.10\,\mathrm{mol}} \]

Explanation: The balanced equation relates two moles of A consumed to four moles of C formed. Therefore, the measured amount of C can be used to calculate the amount of A actually consumed. The relationship is \[ n_{\mathrm{A,consumed}}=\frac24n_{\mathrm{C,formed}} \] However, calculating the fraction of initial A that reacted requires both the consumed amount and the initial amount. The required fraction is \[ f_{\mathrm{A,reacted}}=\frac{n_{\mathrm{A,consumed}}}{n_{\mathrm{A,initial}}} \] The molar mass of C alone does not provide the missing initial amount of A. A balanced chemical equation supplies proportional relationships but does not independently specify how much of each reactant was initially present. The identity of the limiting reagent is also insufficient because the reaction may have stopped before that reagent was exhausted. Knowing which reactant would limit the maximum product amount does not establish the experimentally achieved conversion. By contrast, knowing the original amount of A and measuring the product amount determines both quantities required in the fraction. The measured C amount gives A consumption through the balanced equation. The original A amount provides the denominator. This distinction prevents an observed amount of product from being mistaken automatically for the theoretical maximum. Therefore, Option C provides sufficient information.

Explanation: \(\textbf{Establishing the total initial metal mass:}\) The initial magnesium mass is \[ m_{\mathrm{Mg}}=0.20(24) \] \[ m_{\mathrm{Mg}}=4.8\,\text{g} \] The initial aluminium mass is \[ m_{\mathrm{Al}}=0.10(27) \] \[ m_{\mathrm{Al}}=2.7\,\text{g} \] Therefore, \[ m_{\mathrm{metals,total}}=4.8+2.7 \] \[ m_{\mathrm{metals,total}}=7.5\,\text{g} \] \(\textbf{Determining oxygen incorporated into solids:}\) All \(0.12\,\mathrm{mol}\) oxygen gas is consumed. Since both stated products are solids and no other products form, every consumed oxygen atom becomes part of an oxide. The consumed oxygen mass is \[ m_{\mathrm{O_2,consumed}}=0.12(32) \] \[ m_{\mathrm{O_2,consumed}}=3.84\,\text{g} \] \(\textbf{Calculating the final solid inventory:}\) Initially, all metal is present as solid. The reaction transfers oxygen from the gas phase into solid oxides. No solid material is lost. Therefore, \[ m_{\mathrm{solids,final}}=m_{\mathrm{metals,initial}}+m_{\mathrm{oxygen\ incorporated}} \] \[ m_{\mathrm{solids,final}}=7.5+3.84 \] \[ m_{\mathrm{solids,final}}=11.34\,\text{g} \] \(\textbf{Testing whether complete oxidation of both metals is possible:}\) Complete oxidation of magnesium requires \[ n_{\mathrm{O_2,Mg}}=0.20\left(\frac12\right) \] \[ n_{\mathrm{O_2,Mg}}=0.100\,\mathrm{mol} \] Complete oxidation of aluminium requires \[ n_{\mathrm{O_2,Al}}=0.10\left(\frac34\right) \] \[ n_{\mathrm{O_2,Al}}=0.075\,\mathrm{mol} \] The combined oxygen requirement is \[ 0.100+0.075=0.175\,\mathrm{mol} \] The available \(0.12\,\mathrm{mol}\) is insufficient to oxidise both metals completely. \(\textbf{Assessing product-distribution uniqueness:}\) The problem does not specify how the limited oxygen is distributed between the two metals. The amount of MgO depends on how much magnesium reacts. The amount of \(\mathrm{Al_2O_3}\) depends on how much aluminium reacts. Different distributions can satisfy the same total oxygen consumption while producing different individual oxide masses. \(\textbf{Checking the invariant:}\) Every permitted distribution must incorporate exactly 3.84 g oxygen into the retained solids. The initial 7.5 g of metal atoms also remains in the solids, whether combined or unreacted. Hence, the total final solid mass is fixed even though the separate amounts of the two oxides are not. Assuming preferential reaction of one metal would introduce information not supplied in the question. The justified result is a determined total solid mass with an undetermined individual oxide composition. \(\textbf{Final conclusion:}\) The final solids weigh 11.34 g. Their individual magnesium oxide and aluminium oxide masses cannot be uniquely calculated from the stated information.

Explanation: \(\textbf{Interpreting the formula-unit inventory:}\) The initial solid mixture contains two different ionic compounds. The amount of formula units is the sum of their individual amounts. It does not equal the total amount of ions obtained after dissolution. Let \(x\) be the amount of sodium sulphate and \(y\) the amount of aluminium sulphate. \[ x+y=0.30 \qquad (1) \] \(\textbf{Using sulphate-ion counting:}\) Complete dissociation follows the constituent-ion relationships \[ \mathrm{Na_2SO_4\rightarrow2Na^++SO_4^{2-}} \] \[ \mathrm{Al_2(SO_4)_3\rightarrow2Al^{3+}+3SO_4^{2-}} \] One sodium sulphate formula unit supplies one sulphate ion. One aluminium sulphate formula unit supplies three sulphate ions. Therefore, \[ x+3y=0.50 \qquad (2) \] \(\textbf{Recovering the two salt amounts:}\) Subtracting Equation (1) from Equation (2), \[ 2y=0.50-0.30 \] \[ 2y=0.20 \] \[ y=0.10\,\mathrm{mol} \] Consequently, \[ x=0.30-0.10 \] \[ x=0.20\,\mathrm{mol} \] \(\textbf{Calculating the initial mixture mass:}\) The sodium sulphate molar mass is \[ M_{\mathrm{Na_2SO_4}}=2(23)+32+4(16) \] \[ M_{\mathrm{Na_2SO_4}}=142\,\mathrm{g\,mol^{-1}} \] Its contribution is \[ m_{\mathrm{Na_2SO_4}}=0.20(142) \] \[ m_{\mathrm{Na_2SO_4}}=28.4\,\text{g} \] The aluminium sulphate molar mass is \[ M_{\mathrm{Al_2(SO_4)_3}}=2(27)+3[32+4(16)] \] \[ M_{\mathrm{Al_2(SO_4)_3}}=342\,\mathrm{g\,mol^{-1}} \] Its mass is \[ m_{\mathrm{Al_2(SO_4)_3}}=0.10(342) \] \[ m_{\mathrm{Al_2(SO_4)_3}}=34.2\,\text{g} \] Therefore, \[ m_{\mathrm{mixture}}=28.4+34.2 \] \[ m_{\mathrm{mixture}}=62.6\,\text{g} \] \(\textbf{Counting all independently dispersed ions:}\) Sodium sulphate produces three ions per formula unit. Aluminium sulphate produces five ions per formula unit. \[ n_{\mathrm{ions,total}}=3(0.20)+5(0.10) \] \[ n_{\mathrm{ions,total}}=0.60+0.50 \] \[ n_{\mathrm{ions,total}}=1.10\,\mathrm{mol} \] \(\textbf{Independent elemental-ion check:}\) The individual ion amounts are \[ n_{\mathrm{Na^+}}=2(0.20)=0.40\,\mathrm{mol} \] \[ n_{\mathrm{Al^{3+}}}=2(0.10)=0.20\,\mathrm{mol} \] \[ n_{\mathrm{SO_4^{2-}}}=0.50\,\mathrm{mol} \] Their sum is \(1.10\,\mathrm{mol}\). The positive charge amount is \(0.40+3(0.20)=1.00\,\mathrm{mol}\) of elementary charges. The negative charge amount is \(2(0.50)=1.00\,\mathrm{mol}\) of elementary charges. The solution is electrically neutral. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{mixture}}=62.6\,\text{g},\quad n_{\mathrm{ions,total}}=1.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Distinguishing molecules from constituent atoms:}\) The condition specifies equal total numbers of atoms, not equal numbers of molecules. The three substances have different molecular atomicities. Therefore, the molecular amounts required to satisfy the condition are different. \(\textbf{Determining the atoms per molecule:}\) One ammonia molecule contains one nitrogen atom and three hydrogen atoms. Its atomicity is \[ a_{\mathrm{NH_3}}=1+3=4 \] One methane molecule contains one carbon atom and four hydrogen atoms. \[ a_{\mathrm{CH_4}}=1+4=5 \] One water molecule contains two hydrogen atoms and one oxygen atom. \[ a_{\mathrm{H_2O}}=2+1=3 \] \(\textbf{Connecting molecular amounts with equal atom counts:}\) Let each sample contain \(k\,\mathrm{mol}\) of constituent atoms. For any pure molecular substance, \[ n_{\mathrm{atoms}}=a\,n_{\mathrm{molecules}} \] Therefore, \[ n_{\mathrm{NH_3}}=\frac{k}{4} \] \[ n_{\mathrm{CH_4}}=\frac{k}{5} \] \[ n_{\mathrm{H_2O}}=\frac{k}{3} \] \(\textbf{Converting the molecular amounts into sample masses:}\) The molar masses are \[ M_{\mathrm{NH_3}}=14+3=17\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{CH_4}}=12+4=16\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{H_2O}}=2+16=18\,\mathrm{g\,mol^{-1}} \] The required sample masses are \[ m_{\mathrm{NH_3}}=\frac{17k}{4} \] \[ m_{\mathrm{CH_4}}=\frac{16k}{5} \] \[ m_{\mathrm{H_2O}}=\frac{18k}{3} \] \(\textbf{Constructing the mass ratio:}\) \[ m_{\mathrm{NH_3}}:m_{\mathrm{CH_4}}:m_{\mathrm{H_2O}} =\frac{17}{4}:\frac{16}{5}:6 \] Multiplying every ratio term by 20, \[ m_{\mathrm{NH_3}}:m_{\mathrm{CH_4}}:m_{\mathrm{H_2O}} =85:64:120 \] \(\textbf{Checking the atomic inventory:}\) Choose samples containing four moles of atoms each. The required molecular amounts are \(1\,\mathrm{mol}\) ammonia, \(0.8\,\mathrm{mol}\) methane and \(\frac43\,\mathrm{mol}\) water. Their masses are 17 g, 12.8 g and 24 g, respectively. Multiplication of this mass ratio by five again gives \(85:64:120\). Equal constituent-atom counts do not imply equal molecular counts or equal sample masses. Both atomicity and molar mass must be considered. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{NH_3}}:m_{\mathrm{CH_4}}:m_{\mathrm{H_2O}}=85:64:120} \]

Explanation: \(\textbf{Interpreting the molecular-number ratio:}\) The ratio \(2:1\) means that two methane molecules are present for every one water molecule. The corresponding mole ratio is also \(2:1\). A convenient basis is two moles of methane molecules and one mole of water molecules. \(\textbf{Calculating the component masses:}\) Methane has molar mass \[ M_{\mathrm{CH_4}}=12+4(1)=16\,\mathrm{g\,mol^{-1}} \] The methane mass on the selected basis is \[ m_{\mathrm{CH_4}}=2(16)=32\,\text{g} \] Water has molar mass \[ M_{\mathrm{H_2O}}=2(1)+16=18\,\mathrm{g\,mol^{-1}} \] Its mass is \[ m_{\mathrm{H_2O}}=1(18)=18\,\text{g} \] Therefore, \[ m_{\mathrm{total}}=32+18=50\,\text{g} \] \(\textbf{Determining the water mass percentage:}\) \[ \%\mathrm{H_2O}=\frac{18}{50}\times100 \] \[ \%\mathrm{H_2O}=36\% \] \(\textbf{Counting atoms contributed by methane:}\) One methane molecule contains five atoms. Two moles of methane molecules contain \[ n_{\mathrm{atoms,CH_4}}=2(5) \] \[ n_{\mathrm{atoms,CH_4}}=10\,\mathrm{mol} \] of constituent atoms. \(\textbf{Counting atoms contributed by water:}\) One water molecule contains three atoms. One mole of water molecules therefore contributes \[ n_{\mathrm{atoms,H_2O}}=3\,\mathrm{mol} \] of atoms. The mixture contains \[ n_{\mathrm{atoms,total}}=10+3 \] \[ n_{\mathrm{atoms,total}}=13\,\mathrm{mol} \] of constituent atoms. \(\textbf{Finding the oxygen-atom fraction:}\) Only water contributes oxygen atoms. Every water molecule contains one oxygen atom. Thus, \[ n_{\mathrm{O\ atoms}}=1\,\mathrm{mol} \] Therefore, \[ f_{\mathrm{O,atoms}}=\frac{1}{13} \] \(\textbf{Comparing three different compositions:}\) Water contributes one-third of the mixture's molecules. It contributes 36% of the mixture's mass. Oxygen atoms contribute only one-thirteenth of all constituent atoms. These quantities have different denominators and must not be interchanged. The molecular ratio gives a starting basis, while molar masses and molecular atomicities determine the two requested measures. \(\textbf{Final answer:}\) \[ \boxed{\%\mathrm{H_2O}=36\%,\quad f_{\mathrm{O,atoms}}=\frac1{13}} \]

Explanation: \(\textbf{Translating the molecular inventory:}\) Let \(x\) be the amount of methanol molecules and \(y\) the amount of water molecules. The total molecular amount gives \[ x+y=0.30 \qquad (1) \] \(\textbf{Counting hydrogen atoms in both components:}\) One methanol molecule, \(\mathrm{CH_3OH}\), contains four hydrogen atoms. One water molecule contains two hydrogen atoms. Therefore, the total hydrogen-atom amount satisfies \[ 4x+2y=1.00 \qquad (2) \] \(\textbf{Recovering the component amounts:}\) Multiplying Equation (1) by two, \[ 2x+2y=0.60 \] Subtracting this relation from Equation (2), \[ 2x=0.40 \] \[ x=0.20\,\mathrm{mol} \] Therefore, \[ y=0.30-0.20 \] \[ y=0.10\,\mathrm{mol} \] \(\textbf{Determining methanol mass:}\) The molar mass of methanol is \[ M_{\mathrm{CH_3OH}}=12+4(1)+16 \] \[ M_{\mathrm{CH_3OH}}=32\,\mathrm{g\,mol^{-1}} \] The methanol mass is \[ m_{\mathrm{CH_3OH}}=0.20(32) \] \[ m_{\mathrm{CH_3OH}}=6.4\,\text{g} \] \(\textbf{Recovering the oxygen and carbon inventories:}\) Every methanol molecule contains one oxygen atom and one carbon atom. Every water molecule contains one oxygen atom and no carbon atom. Thus, \[ n_O=x+y \] \[ n_O=0.30\,\mathrm{mol} \] The carbon amount comes entirely from methanol. \[ n_C=x \] \[ n_C=0.20\,\mathrm{mol} \] Therefore, \[ n_O:n_C=0.30:0.20 \] \[ n_O:n_C=3:2 \] \(\textbf{Independent hydrogen check:}\) Methanol contributes \[ n_{H,\mathrm{methanol}}=4(0.20)=0.80\,\mathrm{mol} \] Water contributes \[ n_{H,\mathrm{water}}=2(0.10)=0.20\,\mathrm{mol} \] The combined hydrogen amount is \(1.00\,\mathrm{mol}\), as measured. The calculation requires distinguishing molecules from their constituent atoms and recognising that only one component contributes carbon. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{CH_3OH}}=6.4\,\text{g},\quad n_O:n_C=3:2} \]

Explanation: The sample contains two different diatomic molecular species. Each hydrogen molecule contains two hydrogen atoms. Each oxygen molecule also contains two oxygen atoms. Consequently, every molecule in the mixture contributes exactly two constituent atoms, irrespective of its chemical identity. The measured molecular amount is \(0.40\,\mathrm{mol}\). It therefore necessarily corresponds to \(2(0.40)=0.80\,\mathrm{mol}\) of constituent atoms. The second measurement is consistent with the first, but it provides no new information about the relative amounts of hydrogen and oxygen molecules. Let \(x\) be the amount of oxygen molecules. Then the hydrogen amount is \(0.40-x\). Because both gases are present, \(x\) must be greater than zero and less than \(0.40\,\mathrm{mol}\). The total mass is \[ m=32x+2(0.40-x) \] \[ m=0.80+30x \] The minimum limiting mass of 0.8 g would correspond to pure hydrogen. The maximum limiting mass of 12.8 g would correspond to pure oxygen. Neither limiting composition is allowed because both gases are known to be present. Thus, the total mixture mass lies strictly between these limits, but no unique value can be selected. A measurement of mass, oxygen-atom amount or another composition-sensitive quantity would provide an additional independent constraint. Simply measuring the total number of constituent atoms does not distinguish molecular species having the same atomicity. The correct conclusion is therefore that the composition is underdetermined despite the two mutually consistent particle measurements.

Explanation: \(\textbf{Interpreting complete dissociation:}\) Calcium chloride and aluminium chloride are represented by ionic formula units. Under the stated complete-dissociation assumption, their constituent ions are counted separately in solution. \(\textbf{Calcium chloride contribution:}\) \[ \mathrm{CaCl_2\rightarrow Ca^{2+}+2Cl^-} \] One mole of calcium chloride formula units contributes one mole of calcium ions and two moles of chloride ions. For \(0.10\,\mathrm{mol}\) calcium chloride, \[ n_{\mathrm{Ca^{2+}}}=0.10\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-,from\ CaCl_2}}=2(0.10) \] \[ n_{\mathrm{Cl^-,from\ CaCl_2}}=0.20\,\mathrm{mol} \] \(\textbf{Aluminium chloride contribution:}\) \[ \mathrm{AlCl_3\rightarrow Al^{3+}+3Cl^-} \] For \(0.15\,\mathrm{mol}\) aluminium chloride, \[ n_{\mathrm{Al^{3+}}}=0.15\,\mathrm{mol} \] \[ n_{\mathrm{Cl^-,from\ AlCl_3}}=3(0.15) \] \[ n_{\mathrm{Cl^-,from\ AlCl_3}}=0.45\,\mathrm{mol} \] \(\textbf{Combining chloride ions from both salts:}\) \[ n_{\mathrm{Cl^-,total}}=0.20+0.45 \] \[ n_{\mathrm{Cl^-,total}}=0.65\,\mathrm{mol} \] \(\textbf{Counting all dissolved ions:}\) The solution contains calcium ions, aluminium ions and chloride ions. Therefore, \[ n_{\mathrm{ions,total}}=0.10+0.15+0.65 \] \[ n_{\mathrm{ions,total}}=0.90\,\mathrm{mol} \] \(\textbf{Checking electrical neutrality:}\) The calcium ions contribute positive charge amounting to \[ 2(0.10)=0.20\,\mathrm{mol} \] of elementary charges. The aluminium ions contribute \[ 3(0.15)=0.45\,\mathrm{mol} \] of elementary charges. The combined positive charge amount is \(0.65\,\mathrm{mol}\) of elementary charges. The chloride ions contribute an equal amount of negative charge. The ionic composition is therefore electrically neutral. The amount of formula units originally dissolved is only \(0.25\,\mathrm{mol}\), which must not be confused with the \(0.90\,\mathrm{mol}\) of independently dispersed ions. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{Cl^-}}=0.65\,\mathrm{mol},\quad n_{\mathrm{ions,total}}=0.90\,\mathrm{mol}} \]

Explanation: \(\textbf{Comparing molecular atomicities:}\) A water molecule contains three constituent atoms. A methane molecule contains five constituent atoms. Changing the mixture's molecular composition therefore changes the total number of constituent atoms per mole of gas molecules. \(\textbf{Representing the molecular composition:}\) Let \(x\) be the mole fraction of methane. For the chosen one-mole mixture basis, \[ n_{\mathrm{CH_4}}=x \] \[ n_{\mathrm{H_2O}}=1-x \] \(\textbf{Using the measured total atomic amount:}\) Water contributes three moles of atoms per mole of molecules. Methane contributes five. Therefore, \[ n_{\mathrm{atoms,total}}=3(1-x)+5x \] The measured atomic amount is \(3.60\,\mathrm{mol}\). \[ 3.60=3(1-x)+5x \] \[ 3.60=3-3x+5x \] \[ 3.60=3+2x \] \[ 2x=0.60 \] \[ x=0.30 \] Thus, \[ X_{\mathrm{CH_4}}=0.30 \] and \[ X_{\mathrm{H_2O}}=0.70 \] \(\textbf{Determining the mixture's average molar mass:}\) The molar masses are \[ M_{\mathrm{CH_4}}=16\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{H_2O}}=18\,\mathrm{g\,mol^{-1}} \] For a non-reacting molecular mixture, \[ \overline{M}=X_{\mathrm{CH_4}}M_{\mathrm{CH_4}}+X_{\mathrm{H_2O}}M_{\mathrm{H_2O}} \] Substituting, \[ \overline{M}=0.30(16)+0.70(18) \] \[ \overline{M}=4.8+12.6 \] \[ \overline{M}=17.4\,\mathrm{g\,mol^{-1}} \] \(\textbf{Independent atom-count verification:}\) Methane contributes \[ n_{\mathrm{atoms,CH_4}}=5(0.30)=1.50\,\mathrm{mol} \] Water contributes \[ n_{\mathrm{atoms,H_2O}}=3(0.70)=2.10\,\mathrm{mol} \] Their sum is \(3.60\,\mathrm{mol}\). The total atomic amount identifies the molecular composition because the two molecular species have different atomicities. Once that composition is known, the average molar mass can be calculated from the component molar masses. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{CH_4}}=0.30,\quad\overline{M}=17.4\,\mathrm{g\,mol^{-1}}} \]

Explanation: The comparison concerns the number of nitrogen atoms in equal-mass samples. It does not concern the number of molecules alone. For a substance containing \(a\) nitrogen atoms per molecule, the amount of nitrogen atoms in a sample of mass \(m\) is \[ n_N=\frac{am}{M} \] Since the sample mass is the same for all four substances, nitrogen-atom count is proportional to the quantity \(a/M\). For ammonia, the molar mass is \(14+3=17\,\mathrm{g\,mol^{-1}}\), and there is one nitrogen atom per molecule. Its comparison factor is \[ \frac{a}{M}=\frac1{17} \] For hydrazine, the molar mass is \(2(14)+4=32\,\mathrm{g\,mol^{-1}}\), and there are two nitrogen atoms per molecule. Its comparison factor is \[ \frac{a}{M}=\frac2{32}=\frac1{16} \] For nitrogen dioxide, the molar mass is \(14+2(16)=46\,\mathrm{g\,mol^{-1}}\), and there is one nitrogen atom per molecule. Its factor is \[ \frac1{46} \] For dinitrogen monoxide, the molar mass is \(2(14)+16=44\,\mathrm{g\,mol^{-1}}\), and there are two nitrogen atoms per molecule. Its factor is \[ \frac2{44}=\frac1{22} \] Comparing these positive fractions, \[ \frac1{16}\gt\frac1{17}\gt\frac1{22}\gt\frac1{46} \] Thus, hydrazine supplies the greatest number of nitrogen atoms per gram, followed by ammonia, dinitrogen monoxide and nitrogen dioxide. The molecule containing more nitrogen atoms does not automatically provide more nitrogen atoms per unit mass. Its greater molecular mass must also be considered. The required decreasing order is \(\mathrm{N_2H_4}\gt\mathrm{NH_3}\gt\mathrm{N_2O}\gt\mathrm{NO_2}\).

Explanation: \(\textbf{Distinguishing a phase change from a chemical change:}\) Condensation converts water vapour into liquid water. The chemical formula remains \(\mathrm{H_2O}\). The process does not destroy water molecules or change the numbers of hydrogen and oxygen atoms within them. \(\textbf{Calculating the condensed amount:}\) The initial amount of water molecules is \[ n_{\mathrm{H_2O,initial}}=0.30\,\mathrm{mol} \] Two-thirds condenses. \[ n_{\mathrm{H_2O,liquid}}=\frac23(0.30) \] \[ n_{\mathrm{H_2O,liquid}}=0.20\,\mathrm{mol} \] \(\textbf{Determining the remaining vapour amount:}\) \[ n_{\mathrm{H_2O,vapour}}=0.30-0.20 \] \[ n_{\mathrm{H_2O,vapour}}=0.10\,\mathrm{mol} \] \(\textbf{Checking the total molecular inventory:}\) The two phases contain \[ n_{\mathrm{H_2O,total}}=0.20+0.10 \] \[ n_{\mathrm{H_2O,total}}=0.30\,\mathrm{mol} \] The total amount of water molecules remains unchanged. \(\textbf{Counting oxygen atoms in the vapour only:}\) Every water molecule contains one oxygen atom. Therefore, \[ n_{\mathrm{O,vapour}}=n_{\mathrm{H_2O,vapour}} \] \[ n_{\mathrm{O,vapour}}=0.10\,\mathrm{mol} \] \(\textbf{Independent hydrogen-atom check:}\) The vapour contains \[ n_{\mathrm{H,vapour}}=2(0.10)=0.20\,\mathrm{mol} \] of hydrogen atoms. The liquid contains \[ n_{\mathrm{H,liquid}}=2(0.20)=0.40\,\mathrm{mol} \] of hydrogen atoms. The total hydrogen amount is \(0.60\,\mathrm{mol}\), identical to that in the initial vapour. The phase-specific molecular amount changes, but the total chemical-entity and constituent-atom inventories remain conserved. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{H_2O,total}}=0.30\,\mathrm{mol},\quad n_{\mathrm{O,vapour}}=0.10\,\mathrm{mol}} \]

Explanation: The chemical formula contains two ammonium groups and one sulphate group per formula unit. Expanding the groups gives two nitrogen atoms, eight hydrogen atoms, one sulphur atom and four oxygen atoms. For \(0.10\,\mathrm{mol}\) of formula units, the constituent atomic amounts are proportional to these subscripts. The nitrogen amount is \[ n_N=2(0.10)=0.20\,\mathrm{mol} \] Therefore, report P is correct. The hydrogen amount is \[ n_H=8(0.10)=0.80\,\mathrm{mol} \] Therefore, report Q is correct. The oxygen amount is \[ n_O=4(0.10)=0.40\,\mathrm{mol} \] Therefore, report R is correct. The sulphur amount, omitted from the table's first three reports, is \[ n_S=1(0.10)=0.10\,\mathrm{mol} \] The total constituent-atom amount must include sulphur along with nitrogen, hydrogen and oxygen. Therefore, \[ n_{\mathrm{atoms,total}}=0.20+0.80+0.10+0.40 \] \[ n_{\mathrm{atoms,total}}=1.50\,\mathrm{mol} \] Report S gives only \(1.40\,\mathrm{mol}\), exactly the sum that would result from neglecting sulphur. This omission does not make the three element-specific measurements incorrect. A total atom count must include every element represented in the complete formula. Thus, P, Q and R are consistent, while S is not.