All categories

3 years ago

Are the bases on the interior or the exterior of the double helix? Are they randomly arranged or neatly stacked?

2 answers:

5 0

Each helix corresponds to a nucleotide chain, and the two chains are joined throughout their length by adenine-thymine or guanine-cytosine pairs. These base pairs are STACKED one above the other with their planes perpendicular to the axis of the two spirals. This places the hydrophobic base pairs inside the structure and allows the hydrophilic sugar and phosphate groups to contact water on the exterior. The whole helix will just fit inside a cylinder 2000 pm in diameter.

Lemur [1.5K]3 years ago

3 0

the double helix is hydrogen bonded through the bases only so the bases are inside the helix only

as adenine combines with thymine and guanine with cytosine

phosphate are in the exterior of it

sugar groups constitute the double helix.

You might be interested in

What is the compound NH3

Leona [35]

Ammonia is the compound NH3

4 0

2 years ago

Read 2 more answers

A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

5 0

3 years ago

Yield for this reaction? Reaction: N2(g) + 3 H2(g) → 2 NH3 (g)

andriy [413]

Answer:

Yes, yield.

Explanation:

N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation

First, find limiting reactant:

Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2

Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2

The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

6 0

3 years ago

How many molecules of carbon dioxide, CO2, are present in 388.1 grams?

swat32

Answer:

53.11× 10²³ molecules

Explanation:

Given data:

Number of molecules of CO₂ = ?

Mass of CO₂ = 388.1 g

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of CO₂ = 12× 1 + 16×2

Molar mass of CO₂ = 44 g/mol

Now we will put the values in formula.

Number of moles = 388.1 g/ 44 g/mol

Number of moles = 8.82 moles

Now we will calculate the number of molecules by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

8.82 mol × 6.022 × 10²³ molecules / 1 mol

53.11× 10²³ molecules

6 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

3 years ago