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2 years ago

The sum of all chemical reactions that occur in a cell is called:.

Chemistry

1 answer:

vodomira [7]2 years ago

5 0

Answer:

Metabolism

Explanation:

The sum of the chemical reactions that take place within each cell of a living organism and that provide energy for vital processes and for synthesizing new organic material.

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Explanation:

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3 years ago

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Name at least 4 other gases in the atmosphere besides oxygen and nitrogen

BartSMP [9]

Argon,Carbon dioxide,Neon,Helium, and Methane

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3 years ago

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The colour change of methyl orange with neutral substance

Arisa [49]

Answer:

Pink

Explanation:

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3 years ago

Which of the following does NOT involve a change of state? a. pouring water into a vacuum-insulated bottle b. sublimation of dry

Elena L [17]

<h2> The correct option is (a).</h2>

Explanation:

There is no change of state in pouring water into a vacuum insulated bottle as the vacuum created inside the bottle only reduces the time limit for the transfer of heat by conduction and convection.
It helps the content filled in the bottle to be hot and cool for a longer time.

Hence, option (a) is correct.

Rest of the options (b), (c), (d), (e) is not correct because of the following reasons:

In the sublimation of dry ice, the dry ice (solid-state) is changed to the vapour form.
In the freezing of water, the liquid state is changed to solid-state.
During the vaporization of alcohol, the liquid state is changed to the vapour form.
During the melting of ice, the solid-state is changed to the liquid state.

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3 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago