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1 year ago

The group of molecules called nucleotides contain:

1 answer:

8 0

The group of molecules called nucleotides contain phosphate groups, pyrimidines, purines, and pentose (a 5-carbon sugar). Therefore, E. all of the above is the correct answer.

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Draw a sodium formate molecule. The structure has been supplied here for you to copy. To add formal charges, click the button be

Karo-lina-s [1.5K]

The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.

Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).

Nomenclature:

In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.

Name of Formate:

Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.

E.g.

HCOOH (formic acid) → HCOO⁻ (formate) + H⁺

H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺

Formal Charges:

Formal charges are calculated using following formula,

F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]

For Oxygen:

F.C = [6] - [6 + 2/2]

F.C = [6] - [6 + 1]

F.C = 6 - 7

F.C = -1

For Sodium:

F.C = [1] - [0 + 0/2]

F.C = [1] - [0]

F.C = 1 - 0

F.C = +1

5 0

1 year ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

1 year ago

A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.

Mars2501 [29]

Answer:

The correct answer is "1.0100".

Explanation:

Let the volume of mixture be 100 ml.

then,

The volume of DMSO will be 10 mL as well as that of water will be 90 mL.

DMSO will be:

= $10\times 1.1004$

= $11.004 \ g$

The total mass of mixture will be:

= $90+11.004$

= $101.004 \ g$

Density of mixture will be:

= $\frac{Mass}{Volume}$

= $\frac{101.004}{100}$

= $1.01004 \ g/mL$

hence,

Specific gravity of mixture will be:

= $\frac{Density \ of \ mixture}{Density \ of \ water}$

= $\frac{1.01004}{1}$

= $1.0100$

3 0

2 years ago

7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25

horsena [70]

Answer:

The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.

Explanation:

Mass of compound X = 7.00 g

Moles of compound X = $\frac{7.00 g}{70 g/mol}=0.100 mol$

Mass of water in calorimeter ,m= 35.00 kg = 35000 g

Change in temperature of the water in calorimeter = ΔT

ΔT = 2.113°C

Specific heat capacity of water ,c= 4.186 J/g °C

Q = m × c × ΔT

$Q=35000 g\times 4.186 J/g ^oC\times 2.113^oC=309,575.6 J=309.575 kJ$

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.

Heat of formation of Compound X at 25°C:

$\frac{-Q}{\text{moles of compound X}}=\frac{-309.575 }{0.100 mol}$

= -3095.75 kJ/mol

6 0

1 year ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$