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3 years ago

The process of an atom releasing energy when it moves to a lower energy state is called

Chemistry

1 answer:

Inga [223]3 years ago

4 0

The process of an atom releasing energy when it moves to a lower energy state is called emission.

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g A spontaneous process is one in which: A. releases a large amount of heat B. may happen (is possible) C. will rapidly approach

katrin2010 [14]

Answer:

A. releases a large amount of heat

Explanation:

A reaction is said to be spontaneous if it can proceed on its own without the addition of external energy. A spontaneous reaction is not determined by the length of time, because some spontaneous reactions are completed after a long period of time. They are exothermic in nature. An example is the conversion of graphite to carbon which takes a long period of time to complete. Spontaneous reactions are known to increase entropy in a system. Entropy is the rate of disorder in a system.

In the combustion of fire, energy is released to the surroundings as there is a decrease in energy. This is an example of a spontaneous reaction because it is an exothermic reaction, which causes an increase in entropy and a decrease in energy.

7 0

3 years ago

What element has 4 valence electrons and 5 electron shells?

marta [7]

Answer: Tin (Sn)

Explanation: The electron configuration for tin (Sn) is shown in the picture. It's last electrons are:

5s^2 4d^10 5p^2

The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.

8 0

3 years ago

Which of the following is an example of a consumer?

PIT_PIT [208]

Answer:

D. Rabbit

Explanation:

The rabbit is the consumer because he/she (im not judging) will have to consume other plants or small insects to get his/her (again not judging) energy

5 0

2 years ago

Read 2 more answers

Your winter coat is an example of:

inessss [21]

B. It keeps the warm inside

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2 years ago

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .