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3 years ago

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10.

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

A.

Explanation:

The leading zeros in c, d, and e are not significant. There is no zero in b. Just one left and all it's zeros are significant.

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How many particles are in one mole of copper (II) sulfate, CuSO4?

Kobotan [32]

2. Avogadro’s number

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3 years ago

The monatomic ions of Groups 1A(1) and 7A(17) are all singly charged. In what major way do they differ? Why?

insens350 [35]

<u>Answer:</u> Group 1 ions are known as cations and Group 17 ions are known as anions.

<u>Explanation:</u>

Ions are formed when an atom looses or gains electrons.

If an atom gains electrons, it leads to the formation of negative ions known as anions. <u>For Example:</u> Fluorine is a Group 17 element which gains 1 electron to form $F^-$ ions.

If an atom looses electrons, it leads to the formation of positive ions known as cations. <u>For Example:</u> Sodium is a Group 1 element which looses 1 electron to form $Na^+$ ions.

Hence, group 1 ions are known as cations and Group 17 ions are known as anions.

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3 years ago

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Which statement best describes the compressibility of a gas?

maria [59]

The answer to the question stated above is:
<span> Gas is easily compressible because the molecules of a gas are much further apart than those of a solid.</span>

characteristic properties of gases:

(1) they are easy to compress,

(2) they expand to fill their containers, and

(3) they occupy far more space than the liquids or solids from which they form.

7 0

3 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

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3 years ago

HELLLLLLPPPPPP ME NOW I WILL GIVE YA THE BRAINLIEST ANSWER WHOEVER ANSWERS FIRSTTTTTTT

denis23 [38]

Answer:

I think acetone has more because water is just H20

5 0

3 years ago

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