Answer: The molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Explanation:
Given:
= 43.6 mL,
= 0.125 M
= 25.0 mL,
= ?
Formula used to calculate the concentration of acid is as follows.

Substitute the values into above formula.

Thus, we can conclude that the molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Answer is: Na₂SO₄ (sodium sulfate).
The solubility of sodium sulfate increases up to 32.38°C and than decreases at higher temperatures, the solubility becomes almost independent of temperature.
At 32.38°C it releases of crystal water and melting of the hydrated salt.
Sodium sulfate is the white, inorganic compound.
Sodium sulfate is highly soluble in water.
10g
Explanation:
Box 1, Mass of A = 10g
Box 2, Mass of B = 5g
Box 3, = 1A + 1B
Unknown:
Mass of B that would combine with mass of 20g of A
Solution:
Mass ratio of A to B:
= mass ratio
= mass ratio
The mass ratio of A to B = 2: 1
Now, number of B that will combine with 20g of A;
= mass ratio
= 
Mass of B = 10g
10g of B would combine with 20g of A
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Answer:
see below
Explanation:
The rate constant is missing in question, but use C(final) = C(initial)e^-kt = 0.200M(e^-k·10). Fill in k and compute => remaining concentration of reactant
Answer:
3M
Explanation:
moles ÷ liters = molarity
4.8 ÷ 1.6 = 3M