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4vir4ik [10]
3 years ago
13

If two parents with the genotypes bb and BB have children, what percentage of their children will have bushy eyebrows (dominant)

Chemistry
1 answer:
Leviafan [203]3 years ago
5 0
We should set our possible combinations to determine the phenotypes. 

BB= bushy eyebrows
Bb= bushy eyebrows
bb= no bushy eyebrows

BB x bb

According to the <span>Punnett </span>square that I attached as a pic. There is a 100 percent of all their children to have bushy eyebrows since all will be Bb




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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
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(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

5 0
3 years ago
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