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maxonik [38]
3 years ago
6

Calculate the mass percent of phosphorus in a strand of DNA that consists of equal amounts of each of the four N-bases.

Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

8.62%

Explanation:

The DNA is composed of the grouping of the nucleotides, which are composed of one phosphate group (PO₄⁻²), one molecule of sugar ( deoxyribose), and one N-base, which can be adenine, guanine, cytosine, and thymine.

Let's assume as a calculus basis, 1 mol of the compound, and, because it has the same amount of each N-base, 1 molecule of each N-base nucleotide.

The molar mass of phosphate is 95 g/mol, the molar of the deoxyribose is 134 g/mol, the molar mass of adenine is 135 g/mol, of guanine, is 151 g/mol, of cytosine is 111 g/mol, and of thymine is 126 g/mol.

So, the molar masses of the nucleotide of each N-base are:

Adenine: 95 + 134 + 135 = 364 g/mol

Guanine: 95 + 134 + 151 = 380 g/mol

Cytosine: 95 + 134 + 111 = 340 g/mol

Thymine: 95+ 134 + 126 = 355 g/mol

Thus, the mol with one of each of these N-bases has mass 1439 g (364 + 380 + 340 + 355). The molar mass of phosphorus is 31 g/mol, so 1 mol has 31 g. Each nucleotide has one phosphorus, so the total mass of phosphorus is 124 g (4*31).

The percent of phosphorus is then:

(124/1439)*100% = 8.62%

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Answer:

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Explanation:

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3 years ago
What is the special process called where carbon dioxide and water in the presence of sunlight is turned in carbohydrates
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CO+H2O &Sunlight=O2+C6H6O12
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4 years ago
Brian's aunt has cats. When Brian recently visited her, he started sneezing badly and believes that it was because
kifflom [539]

Answer:

By visiting other households with cats.

Explanation:

This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.

4 0
3 years ago
What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?
PtichkaEL [24]

Answer:

The right choice is CHCl₃

Explanation:

  • <u>To find the empirical formula firstly, change the mass to moles.</u>

For Carbon :

no. of moles = (mass / molar mass) = (5.03 g / 12  g/mol )=  0.42 mol

<u><em>For Hydrogen  :</em></u>

no. of moles = (mass / molar mass) = (0.42 g / 1  g/mol )=  0.42 mol

<u><em>For Chlorine :</em></u>

no. of moles = (mass / molar mass) = (44.5 g / 35.5 g/mol )=  1.25 mol

The ratio for Carbon and Hydrogen is  1 : 1

0.42 mol / 042 mol =1.00

  • <u> Then find a ratio between the moles. </u>

The ratio of Chlorine to both Carbon and Hydrogen is 3:1

1.25 mol / 0.42 mol = 2.98 ≅ 3

So the ratio is 1 C : 1 H : 3 Cl.

So, the right choice is CHCl₃

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Chlorine usually forms one bond to Carbon atom.

Hydrogen usually forms one bond Carbon atom.

4 0
3 years ago
Read 2 more answers
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mandarinka [93]

Answer:

\boxed{\text{-55.8 kJ/mol NaOH}}

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹  

        T₁ = 35.00 °C;          T₂ = 37.00 °C

Calculations:

(a) q₁

n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

 V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

(c) ΔH

0.0300ΔH + 1674 = 0

          0.0300ΔH = -1674

                      ΔH = -1674/0.0300

                      ΔH = -55 800 J/mol

                      ΔH = -55.8 kJ/mol

\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}

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3 years ago
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