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Mrrafil [7]
3 years ago
5

1. Pressure of a gas (decreases, increases) if the volume decreases when temperature

Chemistry
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

If temperature is held constant, the equation is reduced to Boyle's law. Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase. ... Gay-Lussac's law states that at constant volume, the pressure and temperature of a gas are directly proportional.

Explanation:

i hope this helped

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The end point in a titration of a 58mL sample of aqueous HCl was reached by the addition of 25mL of 0.83MNaOH titrant. The react
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Answer:

0.36 M

Explanation:

Let's consider the following neutralization reaction.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

25 mL of 0.83 M NaOH reacted. The moles of NaOH that reacted are:

0.025 L × 0.83 mol/L =0.021 mol

The molar ratio of NaOH to HCl is 1:1. The moles of HCl that reacted are 0.021 moles.

0.021 moles are contained in 58 mL of HCl. The molar concentration of HCl is:

M = 0.021 mol/0.058 L

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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
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Answer:

P=atm

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Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

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