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Nata [24]
3 years ago
15

What is the value of delta Hrxn for this equation:

Chemistry
1 answer:
Ratling [72]3 years ago
6 0

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

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Explanation:

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4 years ago
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Calculate the amount (in grams) of kcl present in 75.0 ml of 2.10 m kcl
Lera25 [3.4K]
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
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C = n/V
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3 years ago
Draw a Lewis structure for the following molecules on scratch paper. Then give the number of bonds between the C and N in each.
Yuliya22 [10]

Refer to the attachments

6 0
3 years ago
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olganol [36]

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Explanation:

Hope it helped!

6 0
3 years ago
How much heat energy would be needed to raise the temperature of a 223 g sample of aluminum [(C=0.895 Jig Cy from 22.5°C to 55 0
dsp73

Answer : The heat energy needed would be, 6486.5125 J

Explanation :

To calculate the change in temperature, we use the equation:

q=mc\Delta T\\\\q=mc(T_2-T_1)

where,

q = heat needed = ?

m = mass of aluminum = 223 g

c = specific heat capacity of aluminum = 0.895J/g^oC

\Delta T = change in temperature

T_1 = initial temperature = 22.5^oC

T_2 = final temperature = 55.0^oC

Putting values in above equation, we get:

q=223g\times 0.895J/g^oC\times (55.0-22.5)^oC

q=6486.5125J

Therefore, the heat energy needed would be, 6486.5125 J

5 0
3 years ago
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