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3 years ago

G what is the difference between the sidechains of leucine and isoleucine? select one:

Chemistry

1 answer:

statuscvo [17]3 years ago

4 0

c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.

The structure of leucine is CH3CH(CH3)CH2CH(NH2)COOH.

The structure of isoleucine is CH3CH2CH(CH3)CH(NH2)COOH.

In leucine, the CH3 group is two carbons away from the α carbon; in isoleucine, the CH3 group is on the carbon next to the α carbon.

Thus, isoleucine has the closer branched carbon.

“One is charged, the other is not” is incorrect. Both compounds are uncharged.

“One has more H-bond acceptors than the other” is incorrect. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.

“They have different numbers of carbon atoms” is incorrect. They each contain six carbon atoms.

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3 years ago

Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr

vredina [299]

Answer : The correct expression for equilibrium constant will be:

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4A+2B\rightleftharpoons 8C$

The expression of $K_c$ will be,

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

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3 years ago

Which change will always take place during nuclear fission?

Aleks04 [339]

Answer:

um pretty sure this is it

Explanation:

The fission process also releases extra neutrons, which can then split additional atoms, resulting in a chain reaction that releases a lot of energy.

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2 years ago

This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press

Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀ = - kt

thus:

(pPH₃ )t = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

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3 years ago