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Ipatiy [6.2K]
2 years ago
14

Mercury(ii) oxide (hgo) decomposes to form mercury (hg) and oxygen (o2). the balanced chemical equation is shown below. 2hgo rig

ht arrow. 2hg o2 the molar mass of o2 is 32.00 g/mol. how many moles of hgo are needed to produce 250.0 g of o2? 3.906 moles 7.813 moles 15.63 moles 73.87 moles
Chemistry
1 answer:
Verizon [17]2 years ago
3 0

Decomposition reaction results in the formation of the products by splitting the reactants. Moles of mercury(ii) oxide needed are 15.63 moles.

<h3>What are moles?</h3>

Moles are the ratio of mass and the molar mass of the compound or the molecule.

Moles of oxygen are calculated as:

\begin{aligned}\rm n &= \rm \dfrac {mass}{molar\; mass}\\\\&= \dfrac{250}{32}\\\\&= 7.8125 \;\rm moles\end{aligned}

The balanced chemical reaction can be shown as,

\rm 2HgO \rightarrow 2Hg + O_{2}

From the reaction, it can be said that 1 mole of oxygen requires 2 moles of mercury oxide.

Moles of mercury oxide are calculated as:

\begin{aligned}  \rm n HgO& = (7.8125 \;\rm moles \; O_{2}) \times (\dfrac{2 \;\rm moles\;  HgO}{1 \;\rm mole \; O_{2}})\\\\\rm n HgO &= 15.625 \;\rm moles \end{aligned}

Therefore, option C. 15.63 moles of mercury (II) oxide is needed.

Learn more about moles here:

brainly.com/question/26898237

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A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express
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Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\  \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are  1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3

We can now calculate the molarity as

Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3}  =13792 \frac{molHNO3}{m3}

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