Question

In: Chemistry

A.) Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is...

A.) Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is burned in a bomb calorimeter, the temperature rises from 24.25 ∘C to 32.33 ∘C.

Calculate the standard enthalpy of formation (ΔH∘f) for nitroglycerin. Express the change in energy in kilojoules per mole to three significant figures.

B.) The air within a piston equipped with a cylinder absorbs 575 J of heat and expands from an initial volume of 0.12 L to a final volume of 0.83 L against an external pressure of 1.0 atm.

What is the change in internal energy of the air within the piston?

C.) A 2.95 g lead weight, initially at 10.8 ∘C, is submerged in 8.07 g of water at 51.9 ∘C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsiu

D.) Suppose that 24 g of each of the following substances is initially at 26.0 ∘C. What is the final temperature of each substance upon absorbing 2.40 kJ of heat?

gold

silver

aluminum

water

Solutions

Expert Solution

A)

we know that

Qcal = Ccal x dT

so

Qcal = 5.86 x ( 32.33 - 24.25)

Qcal = 47.3 kJ

now

Qrxn = -Qcal

so

Qrxn = -47.3 kJ

now

moles of napthalene = 1.025 / 128 = 8.01 x 10-3

so

dErxn = -47.3 / 8.01 x 10-3

dErxn = -5911 kJ / mol

B)

we know that

work done = P(V2-V1)

so

work done = 1 x ( 0.83 - 0.12)

work done = 0.71 L atm

we know that

1 L = 10-3 m3

also

1 atm = 101325 Pa

so

work done = 0.71 x 10-3 x 101325

work done = 71.94 J

now

we know that

Q = dU + W

given

Q = 575

so

575 = dU + 71.94

dU = 503.06

so

the change in internal energy is 503.06 J

C)

we know that

heat lost by hot body = heat gained by cold body

also

heat = m x s x dT

so

ml x sl xd dTl = mw x sw x dTw

let the final temperture be Tf

then

2.95 x 0.128 x (Tf - 10.8) = 8.07 x 4.184 x ( 51.9 - Tf)

Tf = 51.445

so

the final temperature is 51.445 C

D)

we know that

Q = m x s x dT

for gold

2400 = 24 x 0.126 x ( Tf - 26)

Tf = 819.65 C


for silver

2400 = 24 x 0.233 x ( Tf - 26)

Tf = 455.18

for aluminium

2400 = 24 x 0.9 x (Tf - 26)

Tf = 137.11 C

for water

2400 = 24 x 4.184 x (Tf - 26)

Tf = 49.9 C


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