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aleksley [76]
3 years ago
15

Potassium metal reacts with water according to the following balanced equation.

Chemistry
1 answer:
aivan3 [116]3 years ago
8 0

Answer: 1 mole

Explanation:

One mole reacts because there is a 1:1 ratio between the potassium and water at the beginning of the equation. 2K+2H2O has a 1 to 1 ratio. If one potassium reacts there is one h2o that also reacts.

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When a person looks at a bright light, tiny muscles in the eye contract so less light can enter the eye.
Wewaii [24]

Answer:

nonstriated

involuntary

attached to the eyeball

Explanation:

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For the reaction A + B − ⇀ ↽ − C + D A+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing
AURORKA [14]

Answer:

a. Not change the free energy value

b. Increase the free energy value

c. Decrease the free energy value

d. Decrease the free energy value

Explanation:

a. Adding a catalyst:

A catalyst is a substance that will reduce the activation energy of a reaction, it means that the reaction will occur fast. The values of enthalpy, entropy, and free energy are not affected by a catalyst, so ΔG remains the same.

b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

K = ([C]*[D])/([A]*[B])

When [C] and [D] increase, the value of K increases, and lnK also increases, then, the value of ΔG increases.

c. Coupling with ATP hydrolysis:

The free energy can be calculated by:

ΔG = ΔH - TΔS

Where ΔH is the change in enthalpy, and ΔS the change in entropy. The ATP hydrolysis is an exothermic reaction, so ΔH <0. When it is coupled, it will reduce the total value of ΔH, and because of that, the value of ΔG will decrease.

d. Increasing [A] and [B]:

As explained above, the increasing at [A] and [B] will decrease the value of K, so the value of lnK will decrease, and ΔG value will also decrease.

4 0
3 years ago
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Components of crude oil can be separated through a process called _____.
Drupady [299]

Answer:

Components of crude oil can be separated through a process called <u><em>fractional distillation</em></u>.

Explanation:

Fractional distillation can be described as a method of distillation in which a liquid mixture is separated based on the different boiling points of the components present in the mixture. The components of crude oil have different boiling temperatures hence, they can be separated by the process of fractional distillation.

The fractionating column has high temperatures at the bottom so components with high temperatures are separated at the bottom. The upper columns have lower temperatures. Components with lower boiling temperatures get separated at the upper columns.

3 0
3 years ago
The electron configurations of two unknown elements X and Y are shown . X: 1s^ 2 2s^ 2 2p^ 6 3s^ 1 Y: 1s^ 2 2s^ 1 Which statemen
PIT_PIT [208]

Answer:

See below.

Explanation:

They both have 1 valency electron so will be metallic and in the same Group (Group 1) of the Periodic Table, so will have similar properties.

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
3 years ago
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