Electrons orbit the nucleus like planets around the sun.true or false?

Chemistry

2 answers:

zzz [600]3 years ago

8 0

Answer:

True

Explanation:

Someone DELETED my other answer

gogolik [260]3 years ago

7 0

Answer: true

Explanation: Electrons orbit the nucleus of an atom the way that planets revolve around the Sun. The electrons are like the planets in the solar system. The sun is like the nucleus in the solar system. The answer to the question is true.

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2. What is a product?

zysi [14]

Answer:an article or substance that is manufactured or refined for sale.

"food products"

a thing or person that is the result of an action or process.

"her perpetual suntan was the product of a solarium"

Explanation:

6 0

3 years ago

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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.

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