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Ivan
2 years ago
12

What is the outermost layer of the earth's core called?

Chemistry
1 answer:
MatroZZZ [7]2 years ago
6 0
Litter core is that right?
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Which pair of elements form an ionic bond?
Anastasy [175]

Answer:

As an example I can say sodium (Na) and chlorine (Cl).

Explanation:

An ionic bond occurs when a metal element reacts with a nonmetal element. Therefore in the answer given above the Na is metal and Cl is nonmetal and they form a molecule through ionic bonding.

3 0
3 years ago
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Aleks [24]

Answer:

The elements in__Group_ 0 of the Periodic Table are called the_noble__gases. They are generally __unreactive_. because they have a__full_outer shell of electrons. So they do not need to gain__lose_or share _electrons_ with other atoms.

5 0
3 years ago
How the proton H gradient is used to make ATP
sladkih [1.3K]

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The proton gradient produced by proton pumping during the electron transport chain is used to synthesize ATP. Protons flow down their concentration gradient into the matrix through the membrane protein ATP synthase, causing it to spin (like a water wheel) and catalyze conversion of ADP to ATP.

4 0
2 years ago
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4 0
2 years ago
At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091

<u>Initial:</u>         0.225      0.225

<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

K_c=\frac{[ICl]^2}{[Cl_2][I_2]}

Putting values in above equation, we get:

9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0
2 years ago
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