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3 years ago

What are the parts of a nucleotide?

Chemistry

2 answers:

kobusy [5.1K]3 years ago

8 0

B-sugar + phosphate + nitrogen base

alina1380 [7]3 years ago

8 0

Sugar phosphate nitrogen base

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A solution

timurjin [86]

The answer to your question is C. A solution is a homogeneous mixture composed of two or more substances, so it couldn't have been A and D. Since a solution can't have its substances separated by a chemical means because they are chemically bonded, thus they are able to be separated by physical means

3 0

3 years ago

Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct

Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8) = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg 1.12×10¹⁴ g / 1000 = 1.12×10¹¹ kg

6 0

3 years ago

PLEASE HELP ASAP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

In-s [12.5K]

+2, because the element has more protons than electrons now

5 0

3 years ago

A sample of 8.5 g NH 3 on oxidation produces 4.5 g of NO. Calculate the percent yield.

yanalaym [24]

Answer:

nvnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnv

Explanation:

3 0

3 years ago

The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42

denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

$k = Ae^{\frac{-Ea}{RT}}$

<u>Where:</u>

k: is the rate constant

A: is the frequency factor

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

$k_{1} = k_{2}$

$Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}$ (1)

By solving equation (1) for T₁ we have:

$T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C$

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!

4 0

4 years ago