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2 years ago

What is the mass of 0.0150 moles of Na₂SO4 (molar mass = 142.04 g/mol)?

Chemistry

1 answer:

olga2289 [7]2 years ago

4 0

Answer:

2.13 g

Explanation:

(142.04)(0.0150) = 2.13 g

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What are the half-reactions for a galvanic cell with aluminum and gold electrodes?

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3 years ago

What is the ranking order of the degree of polarity? Of HF,HCI,HBr and HI

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The ranking order of the degree of polarity is HF>HCl>HBr>HI

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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.

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3 years ago

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