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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
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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.
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...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
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