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3 years ago

Why is a dominant allele called dominant

Chemistry

2 answers:

Arturiano [62]3 years ago

4 0

Answer:

it dominates the phenotype

Explanation:

GarryVolchara [31]3 years ago

3 0

D, because when you have one recessive and one dominant allele, then the dominant one will always over power the recessive. That's why the majority of people have brown eyes, it's because brown eyes are a dominant trait whereas blue or green eyes are recessive.

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Which of the following would be a value measured using a scale in lab?

Neko [114]

Answer:

A. Actual Yield

Since all of the others are calculated not measured outright.

4 0

3 years ago

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

4 years ago

If 12.7 grams of hydrogen reacted with an unknown amount of oxygen, how many grams of water would be produced? (Write a balanced

nirvana33 [79]

Answer:

48 g.

Explanation:

7 0

2 years ago

Read 2 more answers

For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standa

mart [117]

Answer: The standard enthalpy of formation of this isomer of $C_8H_{18}(g)$ is -210.9 kJ

Explanation:

The given balanced chemical reaction is,

$C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)$

First we have to calculate the enthalpy of formation of $C_8H_{18}$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]$