Answer: it would be a 1 to 1 ratio
Explanation: originally it would be 2 to 2 but you have to reduce
In humans, height, skin color, hair color, and eye color are examples of polygenic traits.
A polygene is a member of a collection of non-epistatic genes that interact additively to steer a phenotypic trait, consequently contributing to more than one-gene inheritance, a sort of non-Mendelian inheritance, in preference to unmarried-gene inheritance, which is the core belief of Mendelian inheritance.
A polygenic trait is a feature, which includes height or skin coloration, that is encouraged by way of or extra genes. because a couple of genes are concerned, polygenic developments do not comply with the styles of Mendelian inheritance. Many polygenic traits are also stimulated by means of the environment and are called multifactorial.
Most inherited trends in animals are polygenic. a few examples are: conformation, kind, size, sturdiness, disorder resistance, temperament, velocity, milk and egg production, growth fee, maturation and sexual adulthood rate, and numerous inherited diseases.
Learn more about polygenic traits here:-brainly.com/question/27493732
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Answer:
what do you need help with i cant see it
Explanation:
As we know that 760 mmHg is equal to 1 atm.
So,
If 760 mmHg is equal to = 1 atm
Then
738 mmHg will be equal to = X atm
Solving for X,
X = (738 mmHg × 1 atm) ÷ 760 mmHg
X = 0.971 atm
Result:
738 mmHg is equal to 0.971 atm.
Answer:
I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)
Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)
a) E(Pt⁺²/Fe°) = - 1.668v
b) Process is Non-spontaneous if E(cell) < 0
Explanation:
Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔
Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)
As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.
E°(Fe⁺²) = -0.44v
E°(Pt⁺²) = +1.20v
E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)
= -0.44v - (+1.20v) = - 1.64v
[Fe⁺²] = 0.0066M
[Pt⁺²] = 0.057M
n = electrons transferred = 2
E(nonstd) = E°(std) - (0.0592/n)logQ);
Q = [Pt⁺²]/[Fe⁺²]
= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v
Also, if ΔG(cell) > 0 => indicates non-spontaneous process
ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)
