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2 years ago

How many elements does hydrochloric acid have

Chemistry

1 answer:

Mars2501 [29]2 years ago

5 0

Answer:

Hydrogen chloride (HCl), a compound of the elements hydrogen and chlorine, a gas at room temperature and pressure. A solution of the gas in water is called hydrochloric acid.

Explanation:

hoped that helped

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Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid

ozzi

Answer:

(a) Hypoeutectic

(b) Alpha solid, aluminium

(d) 97.6% α, 2.4% β

(e) 97.6% α, 2.4% β

(f) 97% α, 3% β

Explanation:

(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic

(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon

% α: Al (11 - 4)/(11 - 1) = 70% α

% L: Si 100 - 70 = 30% β

(d) At 576°C we have

α: 99.83% Si (99.83 - 4)/(99.83- 1.65) = 97.6% α

β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β

(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α

Eutectic 4% Si = 100 - 97.6 = 2.4% β

(f) At 25°C we have;

α%: (99.83 - 4)/(99.83 - 1) = 97% α

β%: 100 - 97 = 3% β.

8 0

3 years ago

solubility of gas solutes in water as temperature increases such as CO2 or O2 gas and explain why using the collision theory.

ludmilkaskok [199]

Answer:

See explanation

Explanation:

When a substance is heated, its average kinetic energy increases as the molecules move faster owing to the supply of energy. The solvent molecules are able to collide more frequently with the solute molecules and dislodge them so that the solute can dissolve in the solvent.

However, when a gaseous solute is dissolved in a liquid; as the temperature is increased and solvent molecules are able to collide more frequently with the solute molecules and dislodge them, gas molecules dissolved in the liquid are more likely to escape to the gas phase and not return due to the increase in their kinetic energy.

Hence, solubility of gas solutes in water decreases as temperature increases.

7 0

2 years ago

Which best describes a chemical reaction that follows the law of conservation of matter?

Leno4ka [110]

I’m thinking C for this hope it’s right

5 0

3 years ago

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41

RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³ M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

(2.14×10⁻³ - x) x x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻]. - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

4 0

3 years ago

A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$