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ivanzaharov [21]
3 years ago
6

Which would exert more pressure on the floor, a 100kg man wearing snow shoes or a 60kg woman wearing high heels? PLEASE EXPLAIN

YOUR ANSWER!
Chemistry
1 answer:
iris [78.8K]3 years ago
5 0

Pressure can be defined as the force acting on a perpendicular surface per unit area.

Force exerted by a man of mass 100 kg wearing snow shoes = m.a

Where m = mass of the man = 100 kg

a = acceleration due to gravity= 9.8 m/s^{2}

Force exerted by the man of mass 100 kg = 100 kg(9.8 m/s^{2}) = 980 N

Force exerted by woman of mass 60 kg = 60 kg(9.8 m/s^{2}) = 588 N

Force exerted by 100 kg man is greater than that exerted as 60 kg woman. The area on which this force is acting determines the pressure. Pressure is inversely proportional to the area on which the force acts. Therefore, the pressure exerted by 100 kg man wearing snow shoes is less than the pressure exerted by a 60 kg woman woman wearing high heels as the force acts over a larger area when the man wears snow shoes when compared to the force exerted over a smaller area in case of the woman wearing high heels.

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Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9

Now, the new equilibrium condition, taking into account the change x, becomes:

9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}

Thus, we arrange the equation as:

\frac{1}{9} (0.9-2x)^2=(0.2+x)^2\\0.09-0.4x+4x^2=0.04+0.4x+x^2\\3x^2-0.8x+0.05=0\\x_1=0.06

Finally, the new concentration is:

[NO]_{eq}=0.9-0.06=0.84M

Best regards.

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