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2 years ago

A solid has an volume

Chemistry

1 answer:

kotykmax [81]2 years ago

7 0

Yes, a solid has a fixed volume.

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If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

2 years ago

1pt Which is an example of an alloy? O A. steel OB. salt water O C. air O D. iron

77julia77 [94]

Steel? Hope this helps?

4 0

3 years ago

What is the definition of a chemical bond?

Gnom [1K]

Answer:

Explanation:

bonding is a process of two different atoms sharing electrons for stability and these electrons are attracted by one atom losing it's electrons to another

7 0

2 years ago

A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti

Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)$

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

$n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2$

In such a way, the yielded moles of hydrobromic acid and chlorine are:

$n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2$

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0

3 years ago

Which of the following best defines the internal energy of a system?

Anarel [89]

In thermodynamics, the internal energy of a thermodynamic system, or a body with well-defined boundaries, denoted by U, or sometimes E, is the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy associated with the vibrational and electric energy of atoms within molecules or crystals. It includes the energy in all the chemical bonds, and the energy of the free, conduction electrons in metals.

7 0

3 years ago