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Iteru [2.4K]
2 years ago
10

Activation process and importance of zymogen

Chemistry
1 answer:
ANEK [815]2 years ago
5 0

Answer:

Zymogen ActivationZymogens are activated by snipping the bonds between two or more amino acids, rather like cutting a balloon string so that it floats away. When the bonds are cut, the enzyme changes its conformation, its 3-D structure, so that the active site is free and able to become active.

Explanation:

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined
Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

Concentration=\frac{number of moles of solute}{Volume}

You know  the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}

The concentrations are:

  • [PCl₃]=\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}
  • [Cl₂]=\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}
  • [PCl₅]=\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}

Replacing:

Q=\frac{0.08*0.16}{0.04}

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

4 0
2 years ago
Which source is most likely to be impartial?
Nina [5.8K]
The answer is A I’m not 100 percent sure tho
6 0
2 years ago
How much is a dogggggggggggggggggg
skelet666 [1.2K]

Answer:

it is 7

Explanation:

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3 0
3 years ago
Read 2 more answers
You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t
solniwko [45]

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

6 0
2 years ago
600.0 mL of air is at 20.0 Celsius. What is the volume of the same air at 60.0 Celsius
REY [17]

Answer:

0.682L or 682mL

Use Charles Law of V1/T1 = V2/T2

V1 = 0.6L

T1 = 293K

V2=?

T2= 333K

Explanation:

?

6 0
2 years ago
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