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taurus [48]
3 years ago
10

What error, if any, would the calculated density of the solid have if the material had a hollow center?

Chemistry
1 answer:
maria [59]3 years ago
3 0

Answer:

3) The calculated density would be high, because the volume would be incorrectly measured low.

Explanation:

Density is defined as mass per unit volume. It means the density is inversely proportional to the volume.

A solid with a hollow center will have high volume due to air inside the hollow area that will lead to low density.

But the error in the calculation of density is that the "density would be high, because the volume would be incorrectly measured low."

Hence, the correct answer is "3)"

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2 nitrogen 4 hydrogen and 3 oxygens 
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Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)
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Answer: Oxygen is the limiting reagent.

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6 0
3 years ago
The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
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When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

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Consider the following reaction at equilibrium: 2CO2 (g) ↔ 2CO (g) + O2 (g) ∆H° = -514 kJ Le Châtelier's principle predicts that
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<h3><u>Answer;</u></h3>

increase the partial pressure of CO2

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