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Kruka [31]
3 years ago
8

1. How many moles are contained in 103.4g of sulfuric acid?

Chemistry
1 answer:
sdas [7]3 years ago
6 0
It is b I hope this helps
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How many liters of octane are in a shipping container that contains 45 moles?
Andrew [12]
D. 1,008 liters because you are looking for liters from a calculation of moles. Recognizing that you can do STP (22.4L) you multiply this number by 45 moles and it is 1,008 liters
3 0
3 years ago
Read 2 more answers
20. What volume of 0.350M KMnO4 solution must be diluted to prepare 600. mL of
Dafna1 [17]

Answer:

25.7 mL

Explanation:

Step 1: Given data

  • Initial volume (V₁): ?
  • Initial concentration (C₁): 0.350 M
  • Final volume (V₂): 600 mL
  • Final concentration (C₂): 0.150 M

Step 2: Calculate the volume of the initial solution

We have a concentrated solution and we want to prepare a diluted one. We can calculate the initial volume using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.150 M × 600 mL / 0.350 M

V₁ = 25.7 mL

3 0
3 years ago
An increase in the temperature of a system at equilibrium favors the
Westkost [7]

Answer:

endothermic reaction.

Explanation:

4 0
3 years ago
Which type of measurement is used more often by scientists?
Alchen [17]
Mass is often the most common and weight is its close to that but I'd go with mass
5 0
3 years ago
The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?
svlad2 [7]

Answer: Dissociation constant of the acid is 10^{-11}.

Explanation: Assuming the acid to be monoprotic, the reaction follows:

                         HA\rightleftharpoons H^++A^-

pH of the solution = 6

and we know that

pH=-log([H^+])

[H^+]=antilog(-pH)

[H^+]=antilog(-6)=10^{-6}M

As HA ionizes into its ions in 1 : 1 ratio, hence

[H^+]=[A^-]=10^{-6}M

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

[HA]=[HA]-[H^+]

[HA]=0.1M-10^{-6}M

[HA]=0.09999M

Dissociation Constant of acid, K_a is given as:

K_a=\frac{[A^-][H^+]}{HA}

Putting values of [H^+],[A^-]\text{ and }[HA] in the above equation, we get

K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}

K_a=1.0001\times 10^{-11}M

Rounding it of to one significant figure, we get

K_a=1.0\times 10^{-11}M\approx 10^{-11}M

6 0
3 years ago
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