4 edited body

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of 0.0400400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickel(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate. Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickel(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate. Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of 0.400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickel(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate. Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

3 grammar, spelling, formatting fixes

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickle(II) nitrate}$$$$\text {nickel(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate.Determine Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickle(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate.Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickel(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate. Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

2 TeX edited. Spellings made right.

...Determine Determine the concentration of the original NaOH solution

theThis was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$nickle(II) nitrate$$$$\text {nickle(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate.Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote fistfirst the equation that decribesdescribes the first part of the reaction :

$$NaOH + HCl + -----> NaCl + H2O$$$$NaOH + HCl \rightarrow NaCl + H2O$$

AfterthatAfter that I wrote the second part of the reaction :

$$NaCl + H2O + Ni(NO_3)_2 -----> Ni(OH)_2 + ? + ?$$$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

...Determine the concentration of the original NaOH solution

the was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$nickle(II) nitrate$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate.Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote fist the equation that decribes the first part of the reaction :

$$NaOH + HCl + -----> NaCl + H2O$$

Afterthat I wrote the second part of the reaction :

$$NaCl + H2O + Ni(NO_3)_2 -----> Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

Determine the concentration of the original NaOH solution

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $$NaOH$$ solution to 90.0 mL of .0400 M $$HCl$$.

The solution was then treated with an excess of $$\text {nickle(II) nitrate}$$, resulting in the formation of 1.06 of $$Ni(OH)_2$$ precipitate.Determine the concentration of the original $$NaOH$$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$$NaOH + HCl \rightarrow NaCl + H2O$$

After that I wrote the second part of the reaction :

$$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

1