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2 years ago

What is the molar concentration of [H3O+] in a cola that has a pH of 3.120?

Chemistry

1 answer:

ELEN [110]2 years ago

8 0

Answer:

[H₃O⁺] = 7.586×10⁻⁴ M

Explanation:

The pH of a solution is a measure of the molar concentration of H₃O⁺ ions in the solution.

The mathematical expresssion that states the relation between the molar concentration of H₃O⁺ ions and the pH of the solution is:

pH = - log [H₃O⁺].

This is pH is numerically equal to the negative decimal logarithm of the molar concentration of H₃O⁺.

The square brackets are used to indicate molar concentration.

Thus:

pH = - log [H₃O⁺] ← equation

3.120 = -log [H₃O⁺] ← substituting values

- 3.120 = log [H₃O⁺] ← product property of the multiplication

$10^{- 3.120}=[H_3O^+]$ ← antilogarithm property

[H₃O⁺] = 7.586×10⁻⁴ M ← result

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An aqueous solution of iron(II) iodide has a concentration of 0.215 molal. The percent by mass of iron(II) iodide in the solutio

Basile [38]

Answer:

6.24%

Explanation:

Molality by definition means a measurement of the number of moles of solute in solution with 1000 gm or 1Kg solvent. Notice the difference that Molarity is defined on the volume of solution and Molality on the mass of solvent.

So, An aqueous solution of iron(II) iodide has a concentration of 0.215 molal.

means 0.215 moles are present in 1 Kg of solvent.

The molar mass of Fe2I = 309.65 g / mole

mass of FeI2 = moles x molar mass

= 0.215 x 309.65

=66.57 gm

mass % of FeI2 = mass of FeI2 x 100 / total mass

= 66.57x 100 / (1000 +66.57)

= 6.24%

8 0

2 years ago

diamond has a density of 3.26 g/cm3. what is the mass of a diamond that has a volume of 0.313 cm cubed

DaniilM [7]

D = m / V

3.26 = m / 0.313

m = 3.26 x 0.313

m = 1.02038 g

hope this helps!

6 0

2 years ago

ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea

kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

Step 1: Data given

H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Step 2: Calculate ΔG

ΔG = ΔG° + RTln Q

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0

3 years ago

Of the following substances, ________ has the highest boiling point. Of the following substances, ________ has the highest boili

kolezko [41]

Answer:

Of the following substances, PCl₃ has the highest boiling point

Explanation:

The given substances and their boiling point from online resources are;

Krypton, Kr, boiling point = -153.4°C

Chlorine gas, Cl₂, boiling point = -34.6°C

Borane, BH₃, (transient compound), boiling point of the dimer it forms B₂H₆ = -164.85°C

Methane, CH₄, boiling point = -161.6°C

Phosphorus trichloride, PCl₃, boiling point = 76.1°C

Therefore, given that PCl₃ is a volatile liquid at room temperature, while the other substances are gases and that PCl₃ has the highest boiling point of all the substances of 76.1°C, the substance with the highest boiling point is PCl₃.

5 0

2 years ago

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of

USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃ = 18.1g

Mass of Cu₂O = 90.4g

Unknown:

Limiting reactant = ?

Mass of N₂ formed = ?

Solution:

The reaction equation is given as:

Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol

Molar mass of NH₃ = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = $\frac{18.1}{143.2}$ = 0.13moles

Number of moles of NH₃ = $\frac{90.4}{17}$ = 5.32moles

From this reaction;

1 mole of Cu₂O combines with 2 mole of NH₃

So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃

= 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

Mass = number of moles x molar mass

1 mole of Cu₂O will produce 1 mole of N₂

0.13 mole of Cu₂O will produce 0.13 mole of N₂

Mass = 0.13 x (2 x 14) = 3.64g

5 0

3 years ago