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2 years ago

How does the flow of energy during freezing compare with the flow of energy during melting

Chemistry

1 answer:

lianna [129]2 years ago

3 0

Answer:

At the melting point, some molecules overcome the forces of attraction. Energy gained after the the solid melts increases the average kinetic energy or the temperature. When liquid is freezing the energy flows out the the liquid. As the kinetic energy decreases, they move more slowly.

Explanation:

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What can you say about the reactants of photosynthesis and the products of cellular respiration?

xz_007 [3.2K]

Answer:

Photosynthesis converts carbon dioxide and water into oxygen and glucose. Glucose is used as food by the plant and oxygen is a by-product. Cellular respiration converts oxygen and glucose into water and carbon dioxide. Water and carbon dioxide are by-products and ATP is energy that is transformed from the process.

Explanation:

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1 year ago

How many moles of HCl are required to neutralize 20.0 mL of 1.5 M KOH?

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2 years ago

The property of a substance stays ______________ no matter how the substance is used.

Irina18 [472]

Any substance changes to another substance that means the change of the physical property. Like water () has different state which changes as the temperature changes. It remain as liquid in the room temperature, in solid form at or below 0°C and vapor phase on or above 100°C. But in all the stage or phase of the substance the composition of the water i.e. remains. Thus the chemical property remains fixed when a substance change to other substance.

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2 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

2 years ago

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

ZanzabumX [31]

The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg.

7 0

2 years ago