PLEASE HELP!!!!!

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni

torisob [31]

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

$ka = \frac{[H_3O^+][X^-]}{[HX]}$

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

$[H_3O^+]} = 10^{-2,40}$

[H₃O⁺] = 0,004M

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = 0,085M

$ka = \frac{[0,004][0,004]}{[0,085]}$

ka = 1,88x10⁻⁴

And pka = 3,73

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

pkb = 10,27

I hope it helps!

4 0

2 years ago

Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon m

weeeeeb [17]

On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ 0,082 L·atm/K·mol · 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.

6 0

2 years ago

Read 2 more answers

`suppose you were tasked with producing some nitrogen monoxide (a.k.a. nitric oxide). i'm sure this is often requested of you. y

Umnica [9.8K]

N(H₂O):n(NO)=6:4(3:2), n(NO)=2·3,5mol÷3=2,33mol
So it is a).

7 0

2 years ago

Read 2 more answers

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

sergeinik [125]

Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.036412 / 15.999 = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$

7 0

2 years ago

Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O

oee [108]

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the number of Cr atoms on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, 6 electrons are gained by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to balance the total charge on the reactant and product side, 14 H⁺ is added on the reactant side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Therefore, the correct balanced reduction half-reaction is:

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

3 0

2 years ago