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2 years ago

What type of reaction occurs when a candle is burned?

Physics

1 answer:

mojhsa [17]2 years ago

8 0

Answer is B - Exothermic because energy is released

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A rigid container holds 0.30g of hydrogen gas.

Strike441 [17]

Answer:

Part A: $\mathbf{Q =94 \ J}$ to two significant figures

Part B: $\mathbf{Q =160 \ J}$ to two significant figures

Part C: $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures

6 0

2 years ago

Could a car drive on a frictionless surface? Explain using the terms action

Julli [10]

Answer:

No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place

7 0

2 years ago

A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.

nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;