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3 years ago

What is the number of moles present?

Chemistry

2 answers:

frosja888 [35]3 years ago

6 0

Answer:

0.056

Explanation:

If your talking about the question

"Determine the number of moles of air present in 1.35 L at 750 torr and 17.0°C. Which equation should you use?" proceeded by "What is the number of moles present?"

The answer is 0.056 moles.

If not sorry :/

Ksenya-84 [330]3 years ago

4 0

this needs more information, did it come with an explination or picture?

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A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a

Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.40 M

$V_1$ = volume of stock solution = 72.0 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = 248 ml

$1.40\times 72.0=m\times 248$

$m=0.41M$

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.41 M

$V_1$ = volume of stock solution = 124 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = (124 +133) ml = 257 ml

$0.41\times 124=m\times 257$

$m=0.20M$

Thus the final concentration of the solution is 0.20 M.

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4 years ago

5. If a truck goes 30 miles in 30 minutes, what is its average speed?

Shtirlitz [24]

Answer:

1 hour

Explanation:

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4 years ago

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What tells the number of protons in the nucleus of every atom of an element

harina [27]

The atomic number is the protons.
Also, the atomic number subtracted by the neutrons are the protons.<span />

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3 years ago

Aqueous acetic acid is neutralized by aqueous potassium hydroxide. True or False

liberstina [14]

Answer:

i think its true but I’m not sure

Explanation:

I know that they can. Be mixed

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3 years ago

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A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH

bearhunter [10]

Answer:

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = 1.36 x 10^-3

Explanation:

The reaction

CICH2COOH ⇄ H+ (aq) + CICH2COO- (aq)

The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.

let us calculate Ka

First , find change in concentration

since , 11% ionized

change in concentration = 0.10 X 11% = 0.011 M

Initial Concentration of CICH2COOH = 0.10 M

change in concentration of CICH2COOH = - 0.011 M

Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

Initial Concentration of CICH2COO- = 0 M

change in concentration of CICH2COO- = + 0.011 M

Equilibrium Concentration of CICH2COO- = 0.011 M

Initial Concentration of H+ = 0 M

change in concentration of H+ = + 0.011 M

Equilibrium Concentration of H+ = 0.011 M

Therefore,

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = [H+][CICH2COO-] /[CICH2COOH]

Ka = (0.011 * 0.011) / (0.089)

Ka = 1.36 x 10^-3

6 0

3 years ago