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dedylja [7]
2 years ago
8

In the peptide GLVIW, the C-terminal end is ________.

Chemistry
1 answer:
zheka24 [161]2 years ago
5 0
The peptide given above is made up of the following amino acids: glycine [G], leucine [L], valine [V], isoleucine [I] and tryptophan [W]. These amino acids are joined together by amide bond to form peptide. Peptides usually have two terminals, the N terminal and the C terminal. For GLVIW, the C terminal end amino acid is tryptophan, that is the last amino acid on the peptide chain. The N terminal amino acid is glycine, that is, the first amino acid on the peptide chain.
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A reaction generates hydrogen gas (H) as a product. The re-
Amiraneli [1.4K]

<u>Answer:</u> The average rate of the reaction is 7.82\times 10^{-3}M/min

<u>Explanation:</u>

To calculate the molarity of hydrogen gas generated, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 3.91\times 10^{-2}mol

Volume of solution = 250 mL = 0.250 L    (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M

Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.

To calculate the average rate of the reaction, we use the equation:

\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}

We are given:

Concentration of hydrogen generated = 0.1564 M

Time taken = 20.0 minutes

Putting values in above equation, we get:

\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min

Hence, the average rate of the reaction is 7.82\times 10^{-3}M/min

6 0
2 years ago
What is a homogeneous ?
USPshnik [31]

Consisting of parts all of the same kind material.

3 0
3 years ago
Read 2 more answers
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
Which of the following compounds will not produce O2 gas when it decomposes? (2 points)
Umnica [9.8K]
LiOH is the right answer.
6 0
2 years ago
How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?
tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄  → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CO₂: 1 mole
  • H₄: 4 moles  
  • CH₄: 1 mole
  • H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles  of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
1 year ago
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