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Svetradugi [14.3K]

3 years ago

9

A scientist uses an accelerator and high energy electrons to study the particles inside the protons of a helium atom. What parti

cles is the scientist studying? A. the He neutrons B. the He nucleus C.the He electrons D.the He quarks

1 answer:

marissa [1.9K]3 years ago

6 0

<span>B. the He nucleus C.the He electrons D.the He quarks</span>

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If heat transfers from the system (solute) to the surroundings (solvent), then AH is

balu736 [363]

Answer:

B) exothermic.

Explanation:

Hello!

In this case, we need to keep in mind that exothermic reactions release heat, so they increase the temperature as the final energy is less than the initial energy; in contrast, endothermic reactions absorb heat, so they decrease the temperature as the final energy is greater than the initial energy.

In such a way, when a dissolution process shows off a negative enthalpy of dissolution, we infer it is an exothermic process due to the aforementioned; therefore, the answer is:

B) exothermic .

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2 years ago

Which best describes a difference between energy transformation in power plants and dams?

gogolik [260]

Answer:

Which best describes a difference between energy transformations in power plants and dams? Only power plants use fossil fuels to transform energy. Only dams use fission to generate thermal energy. ... Only dams use mechanical energy to produce electricity.

Explanation:

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6 0

3 years ago

An atom of boron has an atomic number of 5 and an atomic mass of 11. The atom contains

Liono4ka [1.6K]

Answer:

5 protons, 5 electrons, and 6 neutrons

3 0

3 years ago

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 $-x\;\text{M}$ . $x > 0$ .

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

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

Rewrite as a quadratic equation and solve for $x$ :

$x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)$

$x\approx 0.00111$ .

The pH of a solution depends on its H⁺ concentration.

At equilibrium

$[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}$ .

$\text{pH} = -\log{[\text{H}^{+}]} = 2.79$ .

5 0

2 years ago

Name three ways chemists keep track of amounts of substances

podryga [215]

By measuring them, by monitoring them, or by separating them.

4 0

3 years ago