Which molecule or reaction supplies the energy for polymerization of nucleotides in the process of transcription?
1 answer:
Answer to this is "the phosphate bonds that are present in the nucleotide triphosphate serves as the substrates" in the process of transcription.
Transcription is the process in which DNA is transformed into RNA with the help of an enzyme known as RNA-polymerase. RNA and DNA are the nucleic acids which are formed by linking the nucleotides together.
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Answer:
6. 7870 kg/m³ (3 s.f.)
7. 33.4 g (3 s.f.)
8. 12600 kg/m³ (3 s.f.)
Explanation:
6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.
1 kg= 1000g
1m³= 1 ×10⁶ cm³
Mass of iron bar
= 64.2g
= 64.2 ÷1000 kg
= 0.0642 kg
Volume of iron bar
= 8.16 cm³
= 8.16 ÷ 10⁶
Density of iron bar
= 7870 kg/m³ (3 s.f.)
7.
Mass
= 1.16 ×28.8
= 33.408 g
= 33.4 g (3 s.f.)
8. Volume of brick
= 12 cm³
Mass of brick
= 151 g
= 151 ÷ 1000 kg
= 0.151 kg
Density of brick
= mass ÷ volume
(3 s.f.)
1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x
=
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
Learn more about molarity here:
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