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3 years ago

What are the 3 parts of a nucleotide

2 answers:

7 0

Like DNA, RNA polymers are make up of chains of nucleotides. These have three parts a five carbon ribose sugar. A phosphate molecule . And one of four nitrogenous bases adenine, guanine, cytosine, or uracil

lara [203]3 years ago

5 0

The 3 parts are

1) A five carbon ribose sugar

2) A Phosphate molecule

3) The four nitrogenous bases

I hope that's help !

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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

8 0

3 years ago

2.In a titration, 25.00 cm3 of a solution of hydrochloric acid reacted with 18.40 cm3 of sodium hydroxide solution of concentrat

telo118 [61]

Answer:

0.11mol/dm³

Explanation:

The reaction expression is given as:

HCl + NaOH → NaCl + H₂O

Volume of acid = 25cm³ = 0.025dm³

Volume of base = 18.4cm³ = 0.0184dm³

Concentration of base = 0.15mol/dm³

Solution:

The concentration of hydrochloric acid = ?

To solve this problem, let us first find the number of moles of the base;

Number of moles = concentration x volume

Number of moles = 0.15mol/dm³ x 0.0184dm³ = 0.00276mol

From the balanced reaction equation;

1 mole of NaOH will combine with 1 mole of HCl

Therefore, 0.00276mol of the base will combine with 0.00276mol of HCl

So;

Concentration of acid = $\frac{number of moles }{volume}$ = $\frac{ 0.00276}{0.025}$ = 0.11mol/dm³

4 0

2 years ago

H+(aq) + OH–(aq) → H2O(l) + 55.8 kJ In this reaction there is conservation of

pentagon [3]

In the equation given above, there is conservation of MASS, CHARGE AND ENERGY.
These three parameters are usually conserved during the course of chemical reactions. When any of these parameter experience a reduction during the course of chemical reaction, such loss is always gained by other elements involved in the same reaction, so that at the end of the day, they are not considered as lost.

4 0

3 years ago

During meiosis, only one chromosome from each homologous is passed on to the offspring. What does this help increase?

Helga [31]

This helps increase variation in the offspring.

6 0

3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) = ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 % = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

6. Fórmula empírica: ZnO

3 0

3 years ago