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Keith_Richards [23]
3 years ago
9

What is the mole fraction of acetic acid in a solution containing 2 moles of vinegar and 3 moles of water? A. 2 B. 3 C. 0.4 D. 0

.5
Chemistry
2 answers:
Delvig [45]3 years ago
8 0

Answer - The right answer for the given question is option c, 0.4

Explanation- As we know when a solution’s mole divides the solute’s mole the quotient (or the answer) we get is the mole fraction of substance.

Since in the question’s data solute, given as vinegar and the moles of vinegar are 2 and the data shows the solution is given as water and the moles of water is given as 3. Now we know mole fraction can be calculated by dividing the two. So by dividing we can get the result as 0.4.

9966 [12]3 years ago
3 0

Answer:

C. 0.4.

Explanation:

<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>

<em></em>

- no. of moles of acetic acid = 2, no. of moles of water = 3.

- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.

<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>

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Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete
oee [108]
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
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pH = -log[H⁺] = -log(0,000293 M).
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5 0
3 years ago
What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)
My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

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The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

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Let's convert the moles of produced oxygen, as to find out the theoretical yield.

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To calculate the percentage yield, the formula is

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8 0
3 years ago
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This also means that Chlorine has a high ionization energy or, in simpler terms, it would be difficult to remove an electron from Chlorine.

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