To solve this problem, we need to follow these steps:

1. First, we need to calculate the amount of heat absorbed by the water using the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.

Since the volume of water is given in dm3, we need to convert it to grams. We know that 1 dm3 of water is approximately equal to 1 kg or 1000 g. So, the mass of the water is 1.50 dm3 * 1000 g/dm3 = 1500 g.

The change in temperature (ΔT) is 35.00°C.

So, q = (1500 g)(4.18 J/g°C)(35.00°C) = 219,450 J = 219.45 kJ.

2. Next, we need to calculate the number of moles of compound B that were burned. We know that the heat of combustion of compound B is 2,150 kJ/mol. This means that 2,150 kJ of heat is released when 1 mol of compound B is burned.

So, the number of moles of compound B that were burned is 219.45 kJ ÷ 2,150 kJ/mol = 0.102 moles.

3. Finally, we can calculate the molar mass of compound B. We know that the mass of the compound burned was 50.0 g. So, the molar mass of compound B is 50.0 g ÷ 0.102 moles = 490.20 g/mol.

So, the molar mass of compound B is approximately 490.20 g/mol.

Question

To solve this problem, we need to follow these steps:

1. First, we need to calculate the amount of heat absorbed by the water using the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.

Since the volume of water is given in dm3, we need to convert it to grams. We know that 1 dm3 of water is approximately equal to 1 kg or 1000 g. So, the mass of the water is 1.50 dm3 * 1000 g/dm3 = 1500 g.

The change in temperature (ΔT) is 35.00°C.

So, q = (1500 g)(4.18 J/g°C)(35.00°C) = 219,450 J = 219.45 kJ.

2. Next, we need to calculate the number of moles of compound B that were burned. We know that the heat of combustion of compound B is 2,150 kJ/mol. This means that 2,150 kJ of heat is released when 1 mol of compound B is burned.

So, the number of moles of compound B that were burned is 219.45 kJ ÷ 2,150 kJ/mol = 0.102 moles.

3. Finally, we can calculate the molar mass of compound B. We know that the mass of the compound burned was 50.0 g. So, the molar mass of compound B is 50.0 g ÷ 0.102 moles = 490.20 g/mol.

So, the molar mass of compound B is approximately 490.20 g/mol.

Knowee AI · Accepted Answer

To solve this problem, we need to follow these steps:

1. First, we need to calculate the amount of heat absorbed by the water using the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.

Since the volume of water is given in dm3, we need to convert it to grams. We know that 1 dm3 of water is approximately equal to 1 kg or 1000 g. So, the mass of the water is 1.50 dm3 * 1000 g/dm3 = 1500 g.

The change in temperature (ΔT) is 35.00°C.

So, q = (1500 g)(4.18 J/g°C)(35.00°C) = 219,450 J = 219.45 kJ.

2. Next, we need to calculate the number of moles of compound B that were burned. We know that the heat of combustion of compound B is 2,150 kJ/mol. This means that 2,150 kJ of heat is released when 1 mol of compound B is burned.

So, the number of moles of compound B that were burned is 219.45 kJ ÷ 2,150 kJ/mol = 0.102 moles.

3. Finally, we can calculate the molar mass of compound B. We know that the mass of the compound burned was 50.0 g. So, the molar mass of compound B is 50.0 g ÷ 0.102 moles = 490.20 g/mol.

So, the molar mass of compound B is approximately 490.20 g/mol.

Compound B is burned in a bomb calorimeter that contains 1.50 dm3of water. When Iburned 50.0 grams of compound B in the calorimeter, the temperature rise of thewater in the calorimeter was 35.00C. If the heat of combustion of compound B is2,150 kJ mol-1, what is the molar mass of compound B?

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