How many ions are present in \(0.20\,\mathrm{mol}\) of \( \mathrm{(NH_4)_2SO_4} \)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Ammonium sulfate dissociates in water as:
\( \mathrm{(NH_4)_2SO_4\rightarrow2NH_4^++SO_4^{2-}} \)
Each formula unit produces \(3\) ions in total.
The number of formula units in \(0.20\,\mathrm{mol}\) is:
\(0.20\times6\times10^{23}\)
Therefore, the total number of ions is:
\(0.20\times3\times6\times10^{23}\)
Therefore, the correct answer is (C).
What is the volume, in \( \mathrm{dm^3} \), of ethane gas, \( \mathrm{C_2H_6(g)} \), produced when \(0.25\,\mathrm{dm^3}\) of ethyne, \( \mathrm{C_2H_2(g)} \), reacts with \(0.40\,\mathrm{dm^3}\) of hydrogen gas, \( \mathrm{H_2(g)} \)?
All volumes are measured under the same conditions.
\( \mathrm{C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)} \)
Correct Answer: \( \boxed{\mathrm{A}} \)
At the same temperature and pressure, gas volumes react in the ratio of their stoichiometric coefficients.
From the equation:
\( \mathrm{C_2H_2:H_2:C_2H_6=1:2:1} \)
To react completely with \(0.25\,\mathrm{dm^3}\) of \( \mathrm{C_2H_2} \), the volume of \( \mathrm{H_2} \) required is:
\(0.25\times2=0.50\,\mathrm{dm^3}\)
Only \(0.40\,\mathrm{dm^3}\) of \( \mathrm{H_2} \) is available, so \( \mathrm{H_2} \) is the limiting reactant.
From the \(2:1\) ratio:
\(V(\mathrm{C_2H_6})=\dfrac{0.40}{2}=0.20\,\mathrm{dm^3}\)
Therefore, the correct answer is (A).
\(5.72\,\mathrm{g}\) of \( \mathrm{Na_2CO_3\cdot10H_2O} \) \(\left(M_{\mathrm{r}}=286\,\mathrm{g\,mol^{-1}}\right)\) is dissolved in water to prepare \(0.4\,\mathrm{dm^3}\) of aqueous solution.
What is the concentration of sodium ions, in \( \mathrm{mol\,dm^{-3}} \), in the resulting solution?
Correct Answer: \( \boxed{\mathrm{D}} \)
Moles of \( \mathrm{Na_2CO_3\cdot10H_2O} \):
\(n=\dfrac{5.72}{286}=0.0200\,\mathrm{mol}\)
Each mole of \( \mathrm{Na_2CO_3\cdot10H_2O} \) produces \(2\) mol of \( \mathrm{Na^+} \).
Moles of sodium ions:
\(n(\mathrm{Na^+})=2\times0.0200=0.0400\,\mathrm{mol}\)
Concentration of sodium ions:
\(c=\dfrac{0.0400}{0.4}=0.10\,\mathrm{mol\,dm^{-3}}\)
Therefore, the correct answer is (D).
