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vampirchik [111]
2 years ago
10

Which atom will gain electrons in the reaction, and how many.

Chemistry
1 answer:
Anna35 [415]2 years ago
7 0

Answer:

Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.

Explanation:

yan na answer

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Kryger [21]

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A communicable disease

Explanation:

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8 0
3 years ago
If i increase the [h+] of a solution by 10x, how much will the ph change? and in what direction will ph change?
igomit [66]
For example...
You have solution with [H+] = 0,01M
>>>> pH = -log[H+] = -log0,01 = 2

And you increase the [H+] by 10x ---> 0,01×10 = 0,1M
>>>> pH = -log0,1 = 1

○ pH decrease by 2x
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4 0
3 years ago
Please help me with this I need this now
vodka [1.7K]

Answer:

C.

Explanation:

We get it from the food we eat by mixing it with fluids. Also, you can always look up the answer.

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7 0
2 years ago
Read 2 more answers
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

8 0
3 years ago
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