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3 years ago

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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3.364 g of hydrated barium chloride of BaCL2.xH2O was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of

atroni [7]

Given:

Mass of hydrated barium chloride = 3.364 g

Total volume of barium chloride V(total)= 250 ml

Volume taken for titration V = 10 ml

Volume of AgNO3 consumed = 46.92 ml

Concentration of AgNO3 = 0.0253 M

To determine:

The value of x i.e. the water of hydration in BaCl2

Explanation:

The net ionic equation is-

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Based on the reaction stoichiometry: Equal moles of Ag+ and Cl- combine to form AgCl

Moles of Ag+ consumed = moles of Cl- present

Moles of Ag+ = V(AgNO3) * M(AgNO3) = 0.04692 * 0.0253 = 0.00119moles

Moles of Cl- present = 0.00119 moles

Thus, 0.00119 moles of Cl- are present in 10 ml of the solution

Therefore, number of moles of Cl- in 250 ml would be-

= 0.00119 * 250 /10 = 0.02975 moles of cl-

Now:

2 moles of Cl- are present in 1 mole of BaCl2

Therefore, 0.02975 moles of Cl- correspond to- 0.02975 * 1/2 = 0.01488 moles of BaCl2

Molar mass of BaCl2 = 208.22 g/mol

Thus, mass of BaCl2 = 0.01488 moles * 208.22 g.mol-1 = 3.098 g

Mass of water of hydration = 3.364 - 3.098 = 0.266 g

# moles of water 'x' = .266/18 = 0.015 ≅ 1

Ans: Formula for hydrated barium chloride = BaCl2. 1H2O

7 0

4 years ago

90 POINTSSSS!!!

Katarina [22]

Answer:

$\large \boxed{\text{-2043.96 kJ/mol}}$

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}$

(b)Total enthalpies of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}$

ΔᵣH° is negative, so the reaction is exothermic.

5 0

3 years ago

What is the molarity of a solution made by dissolving 18.9g of ammonium nitrate in enough water to make 855 ml of solution?

Leya [2.2K]

Answer is: the molarity of a solution is 0.276 M.

V(solution) = 855 mL ÷ 1000 mL/L.
V(solution) = 0.855 L.
m(NH₄NO₃) = 18.9 g; mass of ammonium nitrate.
M(NH₄NO₃) = 80.04 g/mol; molar mass of ammonium nitrate.
n(NH₄NO₃) = m(NH₄NO₃) ÷ M(NH₄NO₃).
n(NH₄NO₃) = 18.9 g ÷ 80.04 g/mol.
n(NH₄NO₃) = 0.236 mol; amount of substance.
c(NH₄NO₃) = n(NH₄NO₃) ÷ V(solution).
c(NH₄NO₃) = 0.236 mol ÷ 0.855 L.
c(NH₄NO₃) = 0.276 mol/L; molarity of ammonium nitrate.

7 0

3 years ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

Neporo4naja [7]

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Consider the following serial dilutions.

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is diluted to 250 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM :

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM :

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Therefore, The dilution scheme will not result in a 200μM solution.

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is diluted to 500 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is diluted to 1000 ml

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM :

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Therefore, The dilution scheme will result in a 200μM solution.

7 0

3 years ago

Gas laws describe and predict the behavior of gases without attempting to explain why they happen

Nutka1998 [239]

The gas laws describe and predict the behavior of gases with an explanation and experimental data

So the given statement is False.

2) The volume of gas can be calculated based on Avagadro's law

It states that the volume of a gas is directly proportional or varies with the moles of the gas. Higher the moles more the volume, condition is the pressure and temperature are constants in the two conditions

Thus as here the pressure and temperature of nitrogen gas is kept constant

V α moles

$\frac{V1}{n1}=\frac{V2}{n2}$

Where

V1 = 6 l

n1 = 0.50 mol

V2 = ?

n2 = 0.75 mol

On putting values

V2 = 6 X 0.75 / 0.5 = 9 L

so resulting volume of the gas will be 9L

5 0

3 years ago

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