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nirvana33 [79]
2 years ago
12

A chemistry student needs of acetic acid for an experiment. He has available of a w/w solution of acetic acid in acetone. Calcul

ate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to significant digits.
Chemistry
1 answer:
Volgvan2 years ago
4 0

Answer:

49.4 g Solution

Explanation:

There is some info missing. I think this is the original question.

<em>A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g  available of a 40.5 %  w/w solution of acetic acid in acetone.  </em>

<em> Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.</em>

<em />

We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

400gSolution \times \frac{40.5gSolute}{100gSolution} = 162 g Solute

Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:

20.0gSolute \times \frac{100gSolution}{40.5gSolute} = 49.4g Solution

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12 g Mg x 1 mole Mg / 24 g Mg

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What volume of a 3.0 M stock solution of H2SO4 is needed to prepare 2.8 L of a 1.6 M H2SO4 solution?
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Use M1V1 = M2V2 to solve
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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

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Explanation:

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