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DerKrebs [107]
2 years ago
14

If you assume Hardy-Weinberg equilibrium for this population, what would be the frequencies of the two alleles after 10 generati

ons?
Chemistry
1 answer:
Nitella [24]2 years ago
7 0

B) 2pq

Explanation:

Hardy-Weinberg equilibrium refers to a model which explains the effect of evolution on the gene pool.

The model is based on the assumptions that if no evolutionary force like genetic drift, natural selection and many other will act on the population and therefore the gene pool  (gene frequency and the genotypic frequency) of a populations remains in equilibrium or constant throughout the generations.

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The volume of 3.73 moles of a gas is 78.3 L at a certain temperature and pressure. At the same temperature and pressure, the mol
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Explanation:

) If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? P PV = nRT. 5.6 (12)=460821) T.

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3. What is the gram-formula mass of (NH4)3PO4?
nasty-shy [4]
The gram-formula mass means that the mass of one mole molecule. Then you just need to calculate the elements together. So the gram-formula mass of (NH4)3PO4 is equals to (14+4)*3+31+16*4= 149g/mol. So the answer is (c)
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What is the calculated value of the cell potential at 298K for an
Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an  electrochemical cell with the following reaction, when the H₂  pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and  the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)

<u>Answer:</u> The cell potential of the given electrochemical cell is 0.273 V

<u>Explanation:</u>

For the given chemical equation:

2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)

The half cell reactions for the given equation follows:

<u>Oxidation half reaction:</u> Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V

<u>Reduction half reaction:</u> H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.0-(-0.14)=0.14V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}

where,

E_{cell} = electrode potential of the cell = ?

E^o_{cell} = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

[H^{+}]=1.39M

[Sn^{2+}]=9.35\times 10^{-4}M

p_{H_2}=6.56\times 10^{-2}atm

Putting values in above equation, we get:

E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V

Hence, the cell potential of the given electrochemical cell is 0.273 V

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it combines with the water and H in the atmosphere and creates sulfuric acid thus making the rain acidic

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