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Kazeer [188]
2 years ago
14

Where do you think information is coded in a DNA molecule?

Chemistry
2 answers:
tester [92]2 years ago
6 0

Answer:

In the nucleobases

Explanation:

The series of the nucleobases make the difference!

Check out this: https://en.m.wikipedia.org/wiki/Nucleobase

Solnce55 [7]2 years ago
5 0
Nucleobases here is your answer
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The terms Q and K refer to reaction components at non-equilibrium and equilibrium conditions, respectively. For a forward reacti
daser333 [38]

Answer:

The value of Q must be less than that of K.

Explanation:

The difference of K and Q can be understood with the help of an example as follows

         A ⇄ B

In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]

In this case ratio of  [B] and  [A]  that is

K =  [B] / [A] which is called equilibrium constant.

But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.

Q = [B] / [A] . It is clear that Q < K before equilibrium.

If Q < K , reaction will proceed towards equilibrium or forward reaction will

proceed .

8 0
2 years ago
What is the mass of an object that has a density of 2.3 g/mL And a volume of 343ML
Bogdan [553]

Answer:

788.9g

Explanation:

density=m/v

2.3g/ml=m/343ml

=343×2.3

=788.9g

7 0
3 years ago
One example of matter is​
zaharov [31]

One example of matter could be <em>Light.</em>

6 0
2 years ago
Read 2 more answers
Compare the values of the spin quantum number for two electrons in the same orbital
Leokris [45]
If there are 2 electrons in the same orbital, the spin numbers would be different for both of these 2 electrons. One would have an up spin and the other a down spin.
3 0
3 years ago
Can someone help?? This is really hard.
Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4  /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0
2 years ago
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