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ICE Princess25 [194]

3 years ago

5

--The pressure for a tire is 123 KPa. Calculate the pressure in atmospheres (atm)?

1 answer:

rusak2 [61]3 years ago

4 0

Answer:

0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar

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Juliette [100K]

Full moon I’m pretty sure... good luck!

3 0

2 years ago

Which of the following statements does not describe a chemical property of oxygen:

g100num [7]

I have a feeling that #2 is the one.

5 0

3 years ago

Explain where the water that collects on the outside of a cold drink on a hot day comes from.

krok68 [10]

Answer:

It is because water molecules in the air condensed on to the container of the drink.

Explanation:

The way this works is the water molecules outside are hot and in the gas state, so when they come into contact with the cold side of the container they lose energy due to heat transfer between the molecules and the container, becoming a liquid on the side of the drink.

4 0

2 years ago

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

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

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

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

5 0

3 years ago

How many significant figures are in this number? 22.0 grams

Inga [223]

Answer:

Hey there!

22.0 has three significant figures.

Let me know if this helps :)

6 0

3 years ago