2. When a 1.000 g sample of the rocket fuel hydrazine, N2H4, is burned in a bomb calorimeter which contains 1200g of water, the temperature rises from 24.62C to 28.16C. If the C for the bomb is 840 J/C
Calculate: q reaction for combustion of a one-gram sample
and
q reaction for combustion of one mole of hydrazine in the bomb calorimeter
Well, well, well, looks like we've got ourselves a scientific party! Let's crunch those numbers, shall we?
First, let's calculate q reaction for the combustion of a one-gram sample of hydrazine in the bomb calorimeter. We can use the formula:
q = mass of water × specific heat capacity of water × change in temperature
The mass of water is given as 1200g, the specific heat capacity of water is approximately 4.18 J/g°C, and the change in temperature is the final temperature (28.16°C) minus the initial temperature (24.62°C).
q = 1200g × 4.18 J/g°C × (28.16°C - 24.62°C)
Put on your mathematician hat and solve that equation, and you'll have the answer for q reaction of a one-gram sample.
Now, for the q reaction of one mole of hydrazine, we'll need to use molar mass. The molar mass of hydrazine (N2H4) is roughly 32 g/mol. So, if we burn one mole of hydrazine, we'll have to multiply q by the molar mass, because we've got a whole lot more fuel to burn!
Are you ready to do some more number gymnastics? Calculate q for one mole of hydrazine by multiplying q reaction for a one-gram sample by the molar mass of hydrazine.
There you have it, my scientific friend! Just remember, in the world of chemistry, always keep a sense of humor and a flamethrower on hand. Just kidding about the flamethrower, safety first!
To calculate the q reaction for the combustion of a one-gram sample of hydrazine, we need to use the formula:
q reaction = C x ΔT
where q reaction is the heat released or absorbed during the reaction, C is the heat capacity of the bomb calorimeter (840 J/°C), and ΔT is the change in temperature (in °C).
Given that the temperature rises from 24.62°C to 28.16°C, we can calculate ΔT as:
ΔT = final temperature - initial temperature
ΔT = 28.16°C - 24.62°C
ΔT = 3.54°C
Now, we can calculate q reaction:
q reaction = C x ΔT
q reaction = 840 J/°C x 3.54°C
q reaction = 2973.6 J
Therefore, q reaction for the combustion of a one-gram sample of hydrazine in the bomb calorimeter is 2973.6 J.
To calculate the q reaction for the combustion of one mole of hydrazine in the bomb calorimeter, we need to convert the mass of hydrazine to moles.
The molar mass of hydrazine (N2H4) is 32.04 g/mol.
The number of moles (n) can be calculated using the formula:
n = mass / molar mass
n = 1.000 g / 32.04 g/mol
n = 0.0312 mol (rounded to 4 decimal places)
Now, we can calculate q reaction using the molar quantity:
q reaction = q reaction for one gram sample x n
q reaction for one mole of hydrazine = 2973.6 J x 0.0312 mol
q reaction for one mole of hydrazine = 92.707 J/mol (rounded to 3 decimal places)
Therefore, q reaction for the combustion of one mole of hydrazine in the bomb calorimeter is approximately 92.707 J/mol.
To calculate the q reaction for the combustion of a one-gram sample of hydrazine in the bomb calorimeter, we need to use the formula:
q = C * ΔT
where q represents the heat transferred, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.
In this case, we are given that the heat capacity of the bomb calorimeter (C) is 840 J/°C and the change in temperature (ΔT) is 28.16°C - 24.62°C = 3.54°C.
Now, let's plug these values into the formula:
q = 840 J/°C * 3.54°C
= 2973.6 J
Therefore, the q reaction for the combustion of a one-gram sample of hydrazine is 2973.6 J.
To calculate the q reaction for the combustion of one mole of hydrazine in the bomb calorimeter, we need to know the molar mass of hydrazine (N2H4) and the number of grams of hydrazine in one mole. The molar mass of N2H4 is 2 * (14.01 g/mol) + 4 * (1.01 g/mol) = 32.05 g/mol.
Since we know that the combustion of a one-gram sample of hydrazine releases 2973.6 J of energy, we can calculate the number of grams of hydrazine in one mole by setting up a proportion:
1 g / 2973.6 J = molar mass (g) / 1 mol
Rearranging the equation, we get:
molar mass (g) = 1 mol * 1 g / 2973.6 J
Now, let's substitute the molar mass of hydrazine into the equation:
molar mass (g) = 1 mol * 32.05 g/mol / 2973.6 J
= 0.0108 g/J
Therefore, the q reaction for the combustion of one mole of hydrazine in the bomb calorimeter is:
q = 0.0108 g/J * 2973.6 J
= 32.13 g
Thus, the q reaction for the combustion of one mole of hydrazine in the bomb calorimeter is 32.13 g.
q = mass H2O x specific heat H2O x (delta T) + Ccal*(delta T) for 1 gram (J/g). If i understand the second part of the question correctly,
Calculate J/mol from (J/g)*(molar mass in g/1 mol) = ??
Then substitute ?? (J/mol) into
q(J/mol) = mass H2O x specific heat H2O x (delta T) + Cp(delta T). Solve for delta T. qrxn for 1 mol in that calorimeter = q for water + q for calorimeter.