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vladimir2022 [97]
1 year ago
10

A solution is prepared by adding 16 g of CH3OH (molar mass 32 g) to 90 g of H2O (molar mass 18 g). The mole fraction of CH3OH in

this solution is closest to which of the following?
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
(E) 0.6
Chemistry
1 answer:
andreyandreev [35.5K]1 year ago
6 0

Answer:

The mole fraction of CH3OH in this solution is closest to 0.1.

Explanation:

The mole fraction of CH3OH can be calculated by dividing the number of mole of CH3OH by the sum of all moles present in the solution. In our example, we have 16 g CH3OH and 90 g H2O. Let´s see how many mols of each constituent we have:

1 mol CH3OH = 32 g ⇒ 16 g CH3OH = 0.5 mol.

1 mol H2O = 18 g ⇒ 90 g H20 = 5 mol

Then, the mole fraction is:

X = mol CH3OH / (mol H2O + mol CH3OH)

X= 0.5 mol / (5 mol + 0.5 mol) ≅ 0.1

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If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
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25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
A student has two samples of NaCl, each one from a different source. Assume that the only potential contaminant in each sample i
bija089 [108]

Answer:

The correct option is;

A. Which sample has the higher purity

Explanation:

The information given relate to the presence of two samples of NaCl, from different sources

The only potential contaminant in each of the sources = KCl

The content of the sample = NaCl

The molar mass of NaCl = 58.44 g/mol

The molar mass of KCl = 74.5513 g/mol

Let the number of moles of KCl in the sample = X

For a given mass of NaCl, KCl mixture, we have;

The molar mass of potassium = 39.0983 g/mol

The molar mass of chlorine = 35.453 g/mol

The molar mass of sodium ≈ 23 g/mol

Therefore;

Each mole of KCl, will yield 35.453 g/mol per 74.5513 g/mol of KCl

While each mole of NaCl will yield 35.453 g/mol per 58.44 g/mol of NaCl

Therefore, the pure sodium chloride sample will yield more chlorine per unit mass of sample.

As such if the two samples have the same mass, the sample with the contaminant of KCl will yield less mass of chlorine per unit mass of the sample, from which the student will be able to tell the purity of the solution.

The sample with the higher purity will yield  a higher mass chlorine per unit mass of the sample.

6 0
2 years ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
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Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

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In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
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First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
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Categorize the following materials: bronze alloy, mouse growing an ear on its back, porcelain dentures
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Bronze alloy and porcelain dentures
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