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murzikaleks [220]
3 years ago
7

You are performing an experiment which requires you to add 2.00 grams of benzene to a reaction. Benzene has a density of 0.790 g

/mL. You should add:
a. 2.53 mL
b. 0.395 mL
c. 1.58 mL
d. 2.00 mL
Chemistry
2 answers:
Scorpion4ik [409]3 years ago
8 0

Answer:

a. 2.53 mL

Explanation:

Hello,

In this case, by knowing the density is defined as the ratio of the mass to the volume:

\rho =\frac{m}{V}

Since both the mass and density of benzene are given, we can easily compute the volume by solving for it:

V=\frac{m}{\rho}=\frac{2.00g}{0.790g/mL}\\  \\V=2.53mL

Therefore the answer is a. 2.53 mL .

Best regards.

____ [38]3 years ago
4 0

Answer:

a. 2.53 mL

Explanation:

The density is an intensive property, that is, it does not depend on the system size or the amount of material in the system. At a given temperature, the density of a liquid is constant, so it can be used to identify a substance.

The density (ρ) results from the quotient between the mass (m) of the liquid and its volume (V). We can use this expression to find the volume corresponding to 2.00 grams of benzene (ρ = 0.790 g/mL).

ρ = m / V

V = m / ρ

V = 2.00 g / (0.790 g/mL)

V = 2.53 mL

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Two liquids are analyzed and found to both be 85.7% carbon and 14.3% hydrogen. At 750 mmHg and 150 C, both are gases. At these c
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Answer:

Molecular formula A: C₅H₁₀

Molecular formula B: C₇H₁₄

Explanation:

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H: 14.3% × (1mol / 1.01g) = 14.158 moles H

Mole ratio of H:C is:

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That means in compounds A and B you have 2 hydrogens per atom of carbon and empirical formila is:

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Using PV = nRT, moles of A and B are:

<em>Where P is pressure (750mmHg / 760 = 0.987atm), V is volume (0.8000L), R is gas constant (0.082atmL/molK), and T is temperature (150°C +273.15 = 423.15K)</em>

Moles A and B: n = PV / RT

n = 0.987atm×0.8000L / 0.082atmL/molK×423.15K

n = 0.0228 moles of A and B.

Using the mass of A and B it is possible to find molar mass of each compound:

A = 1.60g / 0.0228mol = 70.31g/mol

B = 2.22g / 0.0228mol = 97.56g/mol

As empirical formula of both compounds is CH₂, (molar mass = 14.03g/mol). Molecular formula of compounds is:

A = 70.31g/mol / 14.03g/mol = 5 → Molecular formula: 5×CH₂ = <em>C₅H₁₀</em>

B = 97.56g/mol / 14.03g/mol = 7 → Molecular formula: 7×CH₂ = <em>C₇H₁₄</em>

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