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3 years ago

A 1300 mL sample of gas with a molar mass of 71.0 g/mol at STP has what density?

Chemistry

1 answer:

Blizzard [7]3 years ago

6 0

Answer:

0.055g/mL

Explanation:

Data obtained from the question include:

Molar Mass of the gass sample = 71g/mol

Volume of the gas sample = 1300 mL

Density =?

The density of a substance is simply mass per unit volume. It is represented mathematically as:

Density = Mass /volume.

With the above equation, we can easily obtain the density of sample of gas as illustrated below:

Density = 71g / 1300 mL

Density = 0.055g/mL

Therefore, the density of the gas sample is 0.055g/mL

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2 years ago

Which of the following is true of mixtures?

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of

Olin [163]

Answer:

Empirical Formula = C₃H₆O₁

Solution:

Data Given:

Mass of Ethyl Butyrate = 3.61 mg = 0.00361 g

Mass of CO₂ = 8.22 mg = 0.00822 g

Mass of H₂O = 3.35 mg = 0.00335 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

%C = (2.277) × (12 ÷ 44) × 100

%C = 2.277 × 0.2727 × 100

%C = 62.09 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

%H = (0.9279) × (2.02 ÷ 18.02) × 100

%H = 0.9279 × 0.1120 × 100

%H = 10.39 %

%O = 100% - (%C + %H)

%O = 100% - (62.09% + 10.39%)

%O = 100% - 72.48%

%O = 27.52 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 62.09 ÷ 12.01

Moles of C = 5.169 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 10.39 ÷ 1.01

Moles of H = 10.287 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 27.52 ÷ 16.0

Moles of O = 1.720 mol

Step 3: Find out mole ratio and simplify it;

C H O

5.169 10.287 1.720

5.169/1.720 10.287/1.720 1.720/1.720

3.00 5.98 1

3 ≈ 6 1

Result:

Empirical Formula = C₃H₆O₁

8 0

3 years ago

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32

ser-zykov [4K]

Answer:

1.94 × 10⁻³

Explanation:

Step 1: Calculate the concentration of H⁺ ions

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M

Step 2: Calculate the acid dissociation constant (Ka) of the acid

For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.

Ka = [H⁺]²/Ca

Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³

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3 years ago