Answer:
w = -531 kJ
1. Work was done by the system.
Explanation:
Step 1: Given data
- Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
- Change in the internal energy of the system (ΔU°): 156 kJ
Step 2: Calculate the work done (w)
We will use the following expression.
ΔU° = q + w
w = ΔU° - q
w = 156 kJ - 687 kJ
w = -531 kJ
By convention, when w < 0, work is done by the system on the surroundings.
A) Temperature does not affect reaction rate
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Question : <span>The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its?
Answer : A, climate
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</span>
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
Number of molecules NH₃ = 4.45 x 10²²
<h3>Further explanation
</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol of NH₃ = n = 0.074
So the number of molecules :
