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julsineya [31]
2 years ago
12

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a s

pecific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
Chemistry
2 answers:
Mama L [17]2 years ago
7 0
Use the formula, Q= mcT

Q= heat
m= mass= 1.900Kg= 1.900 x 10^3 grams
c= specific heat= 3.21 
T= 4.542 K

Q= (1.900 x10^3g)(3.21)(4.542K)= 14.6 Joules.
OleMash [197]2 years ago
6 0

Answer : The amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat absorb = ?

m = mass of calorimeter = 1.900 kg  = 1900 g

c = specific heat of calorimeter = 3.21J/g.K

\Delta T = change in temperature = 4.542 K

Now put all the given value in the above formula, we get:

Q=1900g\times 3.21J/g.K\times 4.542K

Q=27701.658J=27.70kJ

Now we have to calculate the moles of hexane.

\text{Moles of }C_6H_{14}=\frac{Mass of }C_6H_{14}}{Molar mass of }C_6H_{14}=\frac{0.580g}{86.18g/mole}=0.00674mole

Now we have to calculate the heat produced during the combustion of hexane per mole.

\text{Heat produced}=\frac{27.70kJ}{0.00674mole}=4109.79kJ/mole

Therefore, the amount of heat produced during the combustion of hexane per mole is 4109.79 kJ/mole.

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6 0
2 years ago
A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

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8 0
2 years ago
A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
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Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
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