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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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butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
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4 0
2 years ago
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
3 years ago
4. A balloon is filled with 3.0 L of helium at 310 K. The balloon is placed in an oven where the
77julia77 [94]

Answer:

3,29L

Explanation:

3.29L = V2

 

Formula: V1/T1 = V2/T2

--------------------

Given:

V1 = 3.0 L V2 = ?

T1 = 310 K T2 = 340 K

--------------------

Plugin:

(X stands in place of V2 just to make it easier to look at)

[3.0L / 310K = X / 340K]

(3.0L / 310K = 0.01L/K)

0.01L/K = X / 340K

(multiply 340K on both sides, it cancels out on the right)

0.01L/K * 340K = X

(0.01L/K * 340K = 3.29L)

**3.29L = X**

[or]

**3.29L = V2**

4 0
2 years ago
50.0 grams of KCl is dissolved in water to make a 4.00 L
xeze [42]

Answer:0.1677M

Explanation:

Molarity=moles/volume

Number of moles =mass/molar mass

Once you get the number of moles, you apply it to the molarity formula.

8 0
3 years ago
So im confused i don't know what im supposed to do here
FromTheMoon [43]

So I believe you are supposed to take notes based on the guiding questions (the questions on the side).

3 0
3 years ago
Read 2 more answers
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