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enot [183]
3 years ago
10

A chemist must prepare of hydrochloric acid solution with a pH of at . He will do this in three steps: Fill a volumetric flask a

bout halfway with distilled water. Measure out a small volume of concentrated () stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to significant digits.
Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

1.7 mL

Explanation:

<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>

Step 1: Calculate [H⁺] in the dilute solution

We will use the following expresion.

pH = -log [H⁺]

[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M

Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.

Step 2: Calculate the volume of the concentrated HCl solution

We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL

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How many moles of hydrogen are required to react with 4.6 x 10 22 molecules of nitrogen?
V125BC [204]

Answer: 6 moles

Take a look at the balanced chemical equation for this synthesis reaction

N 2(g] + 3 H 2(g] → 2 NH 3(g]

Notice that you have a  1:3  mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.

So, if you have  2 moles of nitrogen taking part in the reaction, you will need

2 moles N 2 ⋅ 3 moles H 2 /1 mole N 2 = 6 moles H 2

8 0
3 years ago
pressure has little effect on the solubility of liquids and solids because they are almost incompressible. True or False
slavikrds [6]

Pressure has little effect on the solubility of liquids and solids because they are almost incompressible True.

Liquids and solids show little change in solubility with changes in pressure. As expected, gases increase in solubility with increasing pressure. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

External pressure has little effect on liquid and solid solubility. In contrast, the solubility of a gas increases as the partial pressure of the gas above the solution increases.

Solubility is a measure of the concentration of dissolved gas particles in a liquid and is a function of gas pressure. Increasing the gas pressure increases the number of collisions and increases the solubility, and decreasing the pressure decreases the solubility.

Learn more about pressure here : brainly.com/question/28012687

#SPJ4

5 0
1 year ago
What is the pH of a weak acid solution that has an [H+] of 2.1 x 10^-6 M?
yaroslaw [1]

Answer:

5.7 pH

Explanation:

To find pH when given [H+], you use this formula: pH = -log[H+]

-log * 2.1e-6 = 5.6778 = 5.7 pH

3 0
3 years ago
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) The Δ G ° ′ of the reaction is − 9.130 kJ ⋅ mol − 1 . Calculate the equil
Liula [17]

Answer:

K = 39.85

ΔG= -6.9 kJ/mol

Explanation:

Step 1: Data given

The ΔG°′ of the reaction is − 9.130 kJ/mol

Temperature = 25.0 °C = 298 K

Body temperature = 37.0 °C = 310K

the concentration of A is 1.9 M

the concentration of B is 0.80 M

Step 2: The reaction

A (aq) ⇌ B (aq)

Step 3:

ΔG° = -RT ln K

⇒with ΔG° = standard Gibbs free energy change = − 9.130 kJ/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 298 K

⇒with K = the equilibrium constant = TO BE DETERMINED

− 9130 J/mol = - 8.314 * 298 * ln K

ln K = 3.685

K = e^3.685

K = 39.85

Step 4: The reaction at body temperature

ΔG= ΔG°  + RT ln [B]/[A]

⇒with ΔG° =  Gibbs free energy change

⇒with ΔG° = standard Gibbs free energy change = − 9130 J/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 310 K

⇒with [A] = 1.9 M

⇒with [B] = 0.80 M

ΔG= -9130 J/mol + 8.314 J/mol*K * 310 K * ln (1.9/0.80)

ΔG= -9130 J/mol + 2229.4J/mol

ΔG=-6900.6 J/mol = -6.9 kJ/mol

6 0
3 years ago
What is the correct name of the compound that is incorrectly named 1-chloro-3-pentyne?
sleet_krkn [62]
No that is why I am asking for help
3 0
3 years ago
Read 2 more answers
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