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Ivahew [28]
3 years ago
9

What is the effect of an increase in the average kinetic energy of an object

Chemistry
1 answer:
timofeeve [1]3 years ago
7 0
The effect of an increase in the average kinetic energy of an object would be an increase in the temperature of the object.  Temperature is directly proportional to the average kinetic energy of the molecules in the object. Hope this answers the question.
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how to determine the net charge of the tripeptide Asp-Gly-Leu at pH 7. Can someone show in details and tricks on how to solve it
Ugo [173]

Answer:

0!

Explanation:

  • You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:            

           

                   -COOH         -NH3              

pH 7------------------------------------------------------              

Asn               -                      +

Gly                -                      +

Leu               -                      +

  • Now that we have our table we'll sketch our peptide's structure:

<em>HN-Asn-Gly-Leu-COOH</em>

This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:

+1 (HN-Asn)

-1 (Leu-COOH)

=0

The net charge is 0!

I hope you find this information useful and interesting! Good luck!

5 0
3 years ago
How many moles of oxygen gas are required to form 55g of carbon monoxide gas
marta [7]

Answer:

8.33 moles CO2 X. 25mol O2. 16mol CO2. = 13.0 moles

6 0
3 years ago
A 4kg bowling ball is held 2.5 meters above your head. A 0.5 kg tennis ball is held at 12 meters above your head. Which one has
Anettt [7]

Answer:

P.E for the 4kg bowling ball held 2.5 meters above head is 100j, while the other is 60j so the 4kg ball has more potential energy

7 0
3 years ago
What does the cardiovascular system do<br> for the human body?
SashulF [63]

Answer: The blood circulatory system (cardiovascular system) delivers nutrients and oxygen to all cells in the body.

Explanation:

5 0
3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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