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lorasvet [3.4K]
3 years ago
13

Calculate ∆E for the following case:

Chemistry
1 answer:
NARA [144]3 years ago
5 0

Answer:

The internal energy(ΔE) of a substance is calculated below:

From the first law of thermodynamics;

ΔE=q+w

Explanation:

<u>ΔE=q+w</u>

here;

+q=endothermic reaction

-q=exothermic reaction

+w=work done on the system

-w=work done by the system

Given:

q=+1.62kJ=1620J

w=-874J

To solve:

the internal energy(ΔE)

We know:

ΔE=q+w

<em>according to the problem;</em>

ΔE=q-w

since;

w=-874J (i.e.)work is done by the system.

ΔE=1620-874

ΔE=+746J

Therefore the internal energy is +746J

i.e. the option is "c"(+746J)

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A 51.9g sample of iron, which has a specific heat capacity of 0.449·J·g?1°C?1, is put into a calorimeter (see sketch at right) t
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Answer:

the initial temperature of the iron sample is Ti = 90,36 °C

Explanation:

Assuming the calorimeter has no heat loss to the surroundings:

Q w + Q iron = 0

Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )  

Assuming Q= m*c*( T- Tir)  

mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0

 Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )

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7 0
3 years ago
Consider the following reaction: NO( g) + SO 3( g) ⇌ NO 2( g) + SO 2( g) A reaction mixture initially contains 0.86 atm NO and 0
DIA [1.3K]

Answer:

The equilibrium pressure of NO2 is 0.084 atm

Explanation:

Step 1: Data given

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.

Kp = 0.0118

Step 2: The balanced equation

NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)

Step 3: The initial pressures

p(NO) = 0.86 atm

p(SO3) = 0.86 atm

p(NO2) = 0 atm

p(SO2) = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2

p(NO) = 0.86 -x atm

p(SO3) = 0.86 -xatm

p(NO2) = x atm

p(SO2) = x atm

Step 5: Define Kp

Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))

Kp = 0.0118 = x²/(0.86 - x)²

X = 0.08427

p(NO) = 0.86 -0.08427 = 0.77573 atm

p(SO3) = 0.86 -0.08427 = 0.77573 atm

p(NO2) = 0.08427 atm

p(SO2) = 0.08427 atm

The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm

5 0
3 years ago
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