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3 years ago

Nucleotide bases bonded to a sugar phosphate backbone make up _____.

1 answer:

5 0

Nucleotide bases bonded to a sugar phosphate backbone make up nucleic acids such as DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Nucleotides have three major parts: sugars, phosphates, and a nitrogenous base. DNA uses four nitrogenous bases: Adenine, Guanine, Cytosine, and Thymine. RNA uses the same bases except for Thymine, which is replaced by Uracil.

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A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your

amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically, Molarity = $\frac{\text{no. of moles}}{\text{Volume in liter}}$

Also, when number of moles are equal in a solution then the formula will be as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

It is given that $M_{1}$ is 8.00 M, $V_{1}$ is 7.00 mL, and $M_{2}$ is 0.80 M.

Hence, calculate the value of $V_{2}$ using above formula as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$8.00 M \times 7.00 mL = 0.80 M \times V_{2}$

$V_{2} = \frac{56 M. mL}{0.80 M}$

= 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

7 0

2 years ago

When aluminum oxidizes in air it forms aluminum oxide (al2o3) if a 51 sheet?

Ede4ka [16]

Answer: mol Al2O3 x 1 mol Al/ 2 mol Al2O3= .25 mol Al The balanced equation tells us that it takes 4 moles of Al to produce 2 moles of Al2O3. 0.50 moles Al2O3 x (4 moles Al / 2 moles Al2O3) = 1.0 moles Al 1.0 moles Al x (27.0 g Al / 1 mole Al) = 27.0 g Al

8 0

3 years ago

How many molecules of CBr4 are in 250 grams of CBr4

Kazeer [188]

Answer:- $4.54*10^2^3$ molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of $CBr_4$ = 12+4(79.9) = 331.6 g per mol

let's make the set up using dimensional analysis:

$250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})$

= $4.54*10^2^3$ molecules

So, there will be $4.54*10^2^3$ molecules in 250 grams of $CBr_4$ .

8 0

2 years ago

What are the processes that lead to the formation of sedimentary rocks? I WILL GIVE BRAINLIEST

Fynjy0 [20]

Answer:

erosion, weathering, dissolution, precipitation, and lithification So,D.weathering and erosion followed by compaction

Explanation:

mark me brainliest!!

7 0

2 years ago

El fluoruro de hidrógeno HF que se utiliza en

Blizzard [7]

Answer:

25.6g de HF son producidos

Explanation:

...¿Cuánto HF es producido?

Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:

Moles CaF2:

Masa molar:

1Ca = 40g/mol

2F = 19*2 = 38g/mol

40+38 = 78g/mol

50g CaF2 * (1mol/78g) = 0.641 moles CaF2

Moles H2SO4:

Masa molar:

2H = 2g/mol

1S = 32g/mol

4O = 64g/mol

98g/mol

100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4

Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.

Moles HF usando la reacción:

0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF

Masa HF:

Masa molar:

1g/mol + 19g/mol = 20g/mol

1.282 moles HF * (20g/mol) =

<h3>25.6g de HF son producidos</h3>

8 0

3 years ago