Answer:
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The graphics in the attachment is part of the question, which was incomplete.
Answer: Fr = 102N and angle of approximately 11°.
Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:
Fx = 70+40-10 = 100
Fy = 40-20 = 20
Now, as the forces form a triangle, the totalforce is:
Fr = 
Fr = 
Fr = ≈ 102N
To determine the angle requested, we use:
arctg H = 
arctg H = 
H = tg 0.2 ≈ 11°.
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules. Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas. The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.
2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride. So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required. Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0

b) To calculate the constant we have the following expression:

Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K

c) The equilibrium pressure of O₂ over M is:
