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anastassius [24]
2 years ago
6

Q4. What are the different ways in which human being use rocks ?​

Chemistry
2 answers:
Sergio [31]2 years ago
5 0
Some people use rocks spiritually, games, and self protection. hope this helps!
Ghella [55]2 years ago
4 0

Answer:

To hit someone on the head

Explanation:

self protection

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Approximately, how many national parks are there in the United States? a. 25 b. 50 c. 120 d. 474
Len [333]
B, I would believe. There are 58, but B is the closest we can get.
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(GIVING BRAINLIEST) do what the following image says.
S_A_V [24]

Answer:

hope it helps.

<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>

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In the oxidation reduction reaction 2FeCl2 + Cl2 = 2FeCl3 what substance is reduced? What substance is oxidized?
marishachu [46]

Answer:

Iron is oxidized while chlorine is reduced.

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reaction:

2FeCl₂  + Cl₂  →  2FeCl₃

in this reaction the oxidation state of iron is increased from +2 to +3. That's why iron get oxidized and it is reducing agent because it reduced the chlorine. The chlorine is reduced from -2 to -3 and it is oxidizing agent because it oxidized the iron.

2Fe⁺²Cl₂⁻²

2Fe⁺³Cl₃⁻³

The iron atom gives it three electrons to three atoms of chlorine and gain positive charge while chlorine atom accept the electron and form anion.

4 0
3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
24 POINTS!!!!!!!!!!
Oliga [24]
Your answer is going to be A, because it was shoved harder, it will go faster

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