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3 years ago

1. A scientist investigating the cell membrane concludes that a particular substance is transported

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

8 0

Answer

The transport of the substance into the cell is reduced or reversed in the absence of ATP. I think........

Explanation:

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3 years ago

Citrate synthase catalyzes the reaction: ????x????????o????c???????????????????? + ????c????????y???? − ????o???? → c???????????

Nady [450]

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA $\rightarrow$ citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a). It is known that , relation between change in free energy ( $\Delta G$ ) of a reaction and equilibrium constant (K) is as follows.

$\Delta G = -RT \times ln K$

where, T = temperature in Kelvin

The given data is as follows.

T = 310 K, $\Delta G = -31.5 kJ /mol = -31500 J/mol$ (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

ln K = $\frac{-(\Delta G)}{RT}$

= $\frac{31500}{8.314 \times 310}$

ln K = 12.22

K = antilog (12.22)

= $2.1 \times 10^{5}$

Therefore, we can conclude that value of equilibrium constant for the given reaction is $2.1 \times 10^{5}$ .

6 0

3 years ago

In one process, 5.95 kg of caf2 is treated with an excess of h2so4 and yields 2.45 kg of hf. calculate the percent yield of hf.

zysi [14]

Answer is: yield of reaction is 80,3%.
Chemical reaction: CaF₂ + H₂SO₄ → CaSO₄ + 2HF.
m(CaF₂) = 5,95 kg · 1000 g/kg = 5950 g.
n(CaF₂) = m(CaF₂) ÷ M(CaF₂).
n(CaF₂) = 5950 g ÷ 78 g/mol.
n(CaF₂) = 76,28 mol.
From chemical reaction: n(CaF₂) : n(HF) = 1 : 2.
n(HF) = 76,28 mol · 2 = 152,56 mol.
m(HF) = 152,56 mol · 20 g/mol.
m(HF) = 3051,2 g ÷ 1000 g/kg = 3,0512 kg.
yield = 2,45 kg kg ÷ 3,0512 kg · 100% = 80,3%.

6 0

3 years ago

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