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CaHeK987 [17]
3 years ago
10

The maximum oxygen uptake is known as the __________.

Chemistry
2 answers:
Tatiana [17]3 years ago
8 0

Answer:maximum oxygen

uptake is as the VO2 max

Explanation:

andrezito [222]3 years ago
4 0
The maximum oxygen uptake is known as the VO2 max.
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Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin
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<u>Answer:</u> The percentage yield of HF is 73.36 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     ....(1)  

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg  = 6250 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:  

\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol

For the given chemical reaction:

CaF_2+H_2SO_4\rightarrow CaSO_4+2HF

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = \frac{2}{1}\times 80.05=160.1mol of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg

To calculate the percentage yield of hydrofluoric acid, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%

Hence, the percentage yield of HF is 73.36 %

4 0
3 years ago
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