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Andrews [41]
4 years ago
14

Which type of bonding would occur between multiple copper atoms?

Chemistry
2 answers:
Ksju [112]4 years ago
8 0
The answer is D metallic
Phantasy [73]4 years ago
5 0
Metallic
Please mark as brainliest
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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
klio [65]

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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3 years ago
If a 50.0 mL sample of gas exerts a pressure of 0.59 atm at 35.0 Celsius. What volume will the gas occupy at STP?
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Explanation:

P1V1/T1 =P2V2/T2

V2 =P1V1T2/P2T1

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V2 =0.03L

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