<u>Answer:</u> The average rate of the reaction is ![7.82\times 10^{-3}M/min](https://tex.z-dn.net/?f=7.82%5Ctimes%2010%5E%7B-3%7DM%2Fmin)
<u>Explanation:</u>
To calculate the molarity of hydrogen gas generated, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Moles of hydrogen gas = ![3.91\times 10^{-2}mol](https://tex.z-dn.net/?f=3.91%5Ctimes%2010%5E%7B-2%7Dmol)
Volume of solution = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
![\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DH_2%3D%5Cfrac%7B3.91%5Ctimes%2010%5E%7B-2%7Dmol%7D%7B0.250L%7D%3D0.1564M)
Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.
To calculate the average rate of the reaction, we use the equation:
![\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%20of%20the%20reaction%7D%3D%5Cfrac%7B%5Ctext%7BConcentration%20of%20hydrogen%20generated%7D%7D%7B%5Ctext%7BTime%20taken%7D%7D)
We are given:
Concentration of hydrogen generated = 0.1564 M
Time taken = 20.0 minutes
Putting values in above equation, we get:
![\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%20of%20the%20reaction%7D%3D%5Cfrac%7B0.1564M%7D%7B20.0min%7D%5C%5C%5C%5C%5Ctext%7BAverage%20rate%20of%20the%20reaction%7D%3D7.82%5Ctimes%2010%5E%7B-3%7DM%2Fmin)
Hence, the average rate of the reaction is ![7.82\times 10^{-3}M/min](https://tex.z-dn.net/?f=7.82%5Ctimes%2010%5E%7B-3%7DM%2Fmin)
The lower case letter represents the recessive allele.
<u>Answer:</u>
0.800 moles of
will be produced when 0.400 moles of
react and are decomposed
<u>Explanation:</u>
The decomposition of
(aq) ------> 2C2H5OH (aq) + 2CO2
¬moles of
= 0.400 moles
Moles of
produced will be calculated by the formula,
Moles of CO2 = 0.400 moles of
![\left(\frac{2 \text { moles of } \mathrm{CO}_{2}}{1 \text { mole of } C_{6} \mathrm{H}_{12} \mathrm{O}_{6}}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B2%20%5Ctext%20%7B%20moles%20of%20%7D%20%5Cmathrm%7BCO%7D_%7B2%7D%7D%7B1%20%5Ctext%20%7B%20mole%20of%20%7D%20C_%7B6%7D%20%5Cmathrm%7BH%7D_%7B12%7D%20%5Cmathrm%7BO%7D_%7B6%7D%7D%5Cright%29)
Substituting the values,
Moles of
=
= 0.400*2 = 0.800 moles
Therefore, 0.800 moles of
will be produced when 0.400 moles of
react and are decomposed