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ira [324]
3 years ago
12

Match the following.

Chemistry
1 answer:
lidiya [134]3 years ago
6 0

Answer:

Knowing others is intelligence; knowing yourself is true wisdom. mastering others is strength; mastering yourself is true power. if you realize that you have enough, you are truly rich, "And the cause of not following your heart, is in spending the rest of your life wishing you had."

Explanation:

I need points so SORRY!

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Why don’t scientists use the Bohr model above to explain charge transfer?
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Why are some people Left-Handed?​
aksik [14]

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because that's how God created them

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7 0
3 years ago
A sample may contain any or all of the following ions: Hg22+, Ba2+, Mn2+. No precipitate formed when an aqueous solution of NaCl
Bess [88]

Answer:

Only Mn⁺² is present.

Explanation:

  • When an aqueous solution of NaCl is added, the Cl⁻ species is introduced. In the presence of Cl⁻, Hg₂²⁺ would precipitate as Hg₂Cl₂.
  • When an aqueous solution of Na₂SO₄ is added, the SO₄⁻² species is introduced. In the presence of SO₄⁻², Ba⁺² would precipitate as BaSO₄.

No precipitate formed when either of these solutions were added, thus <u>the sample does not contain Hg₂⁺² nor Ba⁺²</u>.

  • Under basic conditions Mn⁺² would precipitate as Mn(OH)₂. A precipitate formed once the solution was made basic, so <u>the sample contains Mn⁺²</u>.
3 0
3 years ago
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and
rusak2 [61]

Answer:

4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)

-) Oxidizing agent: NO_3^-_(_a_q_)

-) Reducing agent: Zn_(_s_)

Explanation:

The first step is separate the reaction into the <u>semireactions</u>:

A.Zn~->Zn(OH)_4^-^2

B.NO_3^-~->~NH_3

If we want to balance in <u>basic medium </u>we have to follow the rules:

1. We adjust the oxygen with OH^-

2. We adjust the H with H_2O

3. We adjust the charge with e^-

Lets balance the first semireaction A. :

Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-

Now, lets balance semireaction B:

NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-

Finally, we have to add the two semireactions:

_________________________________________

8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)

2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)

_________________________________________

(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)

(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)

Cancel out the species on both sides:

8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3

Simplifying the equation :

4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3

The Zn_(_s_) is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The NO_3^-_(_a_q_)is<u> reduced</u> therefore is the <u>oxidizing agent</u>.

4 0
4 years ago
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