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oksano4ka [1.4K]
3 years ago
7

An atom from the column 1 can form an iconic compound with another from column 18

Chemistry
1 answer:
Lyrx [107]3 years ago
7 0
The Answer is False





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Read 2 more answers
What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3N
AlladinOne [14]

<u>Answer:</u> The volume of HCl needed is 0.250 L

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

<u>For sodium carbonate:</u>

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol

The chemical equation for the reaction of sodium carbonate and HCl follows:

Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = \frac{2}{1}\times 0.375=0.750mol of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L

Hence, the volume of HCl needed is 0.250 L

8 0
3 years ago
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