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3 years ago

What happens to a glass of water as the average molecular kinetic energy of the water molecules increases?

Chemistry

2 answers:

Tresset [83]3 years ago

7 0

Answer: The water becomes warmer.

Explanation:

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. The molecules start moving more randomly. It is determined by the equation:

$K=\frac{3RT}{2}$

From above, it is visible that kinetic energy is directly related to the temperature of the system. So, if the average molecular kinetic energy increases, the temperature of the water must have increased and thus the water will become warmer.

trasher [3.6K]3 years ago

4 0

"The water becomes warmer."
<u>Remember</u>: Kinetic energy means how much, on average, a molecule is moving around. This is directly translated into heat. Therefore, the higher the kinetic energy, the more heat produced.

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Explanation:

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3 years ago

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3 years ago

A student places one piece of chalk into a bowl with vinegar and another into a bowl with water. The next day she observes that

USPshnik [31]

Heavy rainfall causes a landslide. Hence, option A is correct.

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2 years ago

Why does a hot air balloon rise when the air inside it is heated

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3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

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3 years ago