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liubo4ka [24]
3 years ago
10

0.47 moles MgCO3 to g. 450.2 g LiNO3 to moles 36.8g Br2 to molecules

Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

- 39.6 g of MgCO₃

- 6.53 moles of LiNO₃

- 1.38×10²³ molecules of Br₂

Explanation:

Let's make some conversions here:

- 0.47 moles of MgCO₃ to g.

We should know, the molar mass of carbonate: 84.3 g /mol

0.47 mol . 84.3 g / 1mol = 39.6 g

- 450.2 g of LiNO₃

Molar mass of nitrate: 68.94 g/mol

450.2 g . 1 mol /68.94g = 6.53 moles of LiNO₃

- Molar mass for dyatomic bromine: 159.80 g/mol

36.8 g . 1 mol/159.80 g = 0.230 moles

1 mol has 6.02×10²³ molecules

0.230 mol . 6.02×10²³ molecules / 1 mol = 1.38×10²³ molecules of Br₂

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4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an averag
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Answer:

C. 0.191 M

Explanation:

Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the <u>reaction between HCl and NaOH</u>:

HCl~+~NaOH~->~NaCl~+~H_2O

The "<u>titrant"</u> in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

0.100~M=\frac{mol}{0.0382~L}

mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH

Now, in the reaction, we have a <u>1:1 molar ratio</u> between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:

0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl

If we want to calculate the molarity of the HCl solution we have to <u>divide by the litters</u> of HCl used in the experiment (20 mL= 0.02 L):

\frac{0.0382~mol~HCl}{0.02~L}~=~0.191~M

The concentration of the HCl solution is 0.191 M

I hope it helps!

8 0
3 years ago
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