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malfutka [58]
3 years ago
15

In any sample of DNA, the amount of adenine (A) equals the amount of _____.

Chemistry
2 answers:
aivan3 [116]3 years ago
8 0
It is equal to amount of thymine
enyata [817]3 years ago
5 0

<u>Answer:</u> The correct answer is thymine.

<u>Explanation:</u>

DNA is known as deoxy ribonucleic acid. This molecule contains 4 nitrogenous bases, which are adenine (A), thymine (T), cytosine (C) and guanine(G). This molecule forms double helix structure.

Adenine formed two hydrogen bonds only with thymine and cytosine forms three hydrogen bonds only with guanine.

Thus, the amount of adenine (A) will be equal to the amount of thymine (T).

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
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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

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Chlorine has a atomic number of 17. It often forms an ion by gaining 1 electron. What would it's charge be?
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Answer:

The charge would be -1.

Explanation:

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