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3 years ago

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During one researcher's experiment, the total mass of the gas-generating solid and entire assembly was 52.1487g before the react

ion and 52.1098g after the reaction. If the number of moles of gas generated was calculated to be 8.854 x 10-4 mol, what is the molar mass of the gas?

1 answer:

irinina [24]3 years ago

7 0

Answer:

43.93 g/mol

Explanation:

The mass of the gas before reaction = 52.1487 g

The mass of the gas after reaction = 52.1098 g

Mass of gas generated = 0.0389 g

Moles of the gas = $8.854\times 10^{-4}\ moles$

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Molar\ mass = \frac{Mass\ taken}{Moles}$

$Molar\ mass= \frac{0.0389\ g}{8.854\times 10^{-4}\ moles}$

Molar mass of the gas = 43.93 g/mol

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20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

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Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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<u>Answer:</u> The net ionic equations are written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

<u>For 1:</u>

The chemical equation for the reaction of calcium nitrate and sodium sulfate is given as:

$Ca(NO_3)_2(aq.)+Na_2SO_4(aq.)\rightarrow CaSO_4(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ca^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$

<u>For 2:</u>

The chemical equation for the reaction of lead (II) nitrate and potassium chloride is given as:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)$

<u>For 3:</u>

The chemical equation for the reaction of calcium chloride and sodium phosphate is given as:

$3CaCl_2(aq.)+2(NH_4)_3PO_4(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4Cl(aq.)$

Ionic form of the above equation follows:

$3Ca^{2+}(aq.)+6Cl^-(aq.)+6NH_4^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4^+(aq.)+6Cl^-(aq.)$

As, ammonium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$3Ca^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)$

<u>For 4:</u>

The chemical equation for the reaction of barium chloride and sodium sulfate is given as:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the net ionic equations are written above.

3 0

3 years ago