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galben [10]
3 years ago
12

Lance is a 60-kg runner who, on a good day, runs at a speed of 5 m/s. What is his kinetic energy when he runs at this speed? *

Chemistry
1 answer:
Contact [7]3 years ago
7 0

Answer:

His kinetic energy when he runs at that 5 m/s is 750 joules

Explanation:

The given information are;

The mass of, Lance, the runner = 60-kg

The speed by which Lance is running = 5 m/s

The kinetic energy, KE, is given by the relation;

KE = 1/2×m×v²

Where;

m = The mass = 60-kg

v = The instantaneous speed of motion = 5 m/s

The kinetic energy can be found by substituting the values of the variables in the kinetic energy equation as follows;

KE = 1/2 × 60 kg × (5 m/s)² = 750 joules

His kinetic energy when he runs at that 5 m/s is 750 joules.

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Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

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Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

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