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3 years ago

12

Which of these changes would cause a decrease in the pressure of a contained gas?

1 answer:

Harman [31]3 years ago

4 0

The equation is pv=nrt. So to decrease pressure, one would increase volume, decrease moles, or decrease temperature.

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Angelina_Jolie [31]

Hello there.

Thomson's atomic model is best described by which of the following statements?

A nucleus with electrons moving around it like planets.

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3 years ago

Which of the following methods can be used to completely separate a solution containing alcohol and water?

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3 years ago

How many electrons will be in the valence of non- H atoms when they make covalent bonds?

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2 years ago

What is the name of this hydrocarbon? <br><br> hexane<br> heptane<br> octane<br> pentane

pav-90 [236]

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3 years ago

Read 2 more answers

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

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3 years ago