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2 years ago

How many photons of light are emitted when the electrons in 10 hydrogen atoms drop from energy level 5 to 3?

Chemistry

1 answer:

nata0808 [166]2 years ago

6 0

The Answer is 10 it a little difficult to explain but I’ll try my best

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Tell me What is an element?

Anarel [89]

Answer:

Any substance that cannot be decomposed into simpler substances by ordinary chemical processes.

Another definition:

An element is the simplest pure substance which can neither be split nor built up from other simpler substances by chemical reaction

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2 years ago

Which of the following is considered a greenhouse gas?

Sergeeva-Olga [200]

The primary greenhouse gases are water vapor, carbon dioxide, methane, nitrous oxide, and ozone. So, the correct answer among these options is water vapor, that is considered to be a greenhouse gas.

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2 years ago

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Ksju [112]

I think A. is the answer

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3 years ago

The quantity, 1,385 mg is equivalent to in scientific notation

Agata [3.3K]

Answer:

The answer is 1385 x10∧ -3

Explanation:

Scientific notation is a quick way to represent a number using notation in exponential form or powers of base ten.

This notation allows us to express too large or small numbers easily.

<u>For example: </u>

500 is written as 5x10∧2; Where the power 2 represents the number of 0's that follow 5.

0.0093 is written as 9.3x10∧-3; Where the power -3 represents the number of times the comma was moved to the right.

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2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago