Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:
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<em>N₂(g) +O₂(g) ⇄ 2 NO(g)
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<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
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Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
![K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}](https://tex.z-dn.net/?f=K%3D7.70%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%20%3D%5Cfrac%7B%282.8-2x%29%5E%7B2%7D%20%7D%7B%281.2%2Bx%29.x%7D)
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M
Ethylene is the starting material for the preparation of a number of two-carbon compounds including ethanol (industrial alcohol), ethylene oxide (converted to ethylene glycol for antifreeze and polyester fibres and films), acetaldehyde (converted to acetic acid), and vinyl chloride (converted to polyvinyl chloride).
There are three types of
wind load or wind forces.
<span>Uplift load is the first one;
this is when the wind flow pressures create a strong lifting effect. Wind flow under the roof pushes upward, while
wind flow over the roof goes upward.</span>
Answer:
There are often not more than one or two independent variables tested in an experiment.
Answer:
162.2 g/mol
Explanation:
Molar mass is defined as the mass in 1 mole of the substance. It is calculated by adding the molar mass of the substituents each multiplied by the subscript they have in the formula.
Thus, To calculate molar mass of nicotine having formula, 
Molar mass of C = 12.0107 g/mol
Molar mass of H = 1.00784 g/mol
Molar mass of N = 14.0067 g/mol
Thus, molar mass of nicotine = 
= 162.2 g/mol
