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Answer:
- <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>
Explanation:
Assume 100% ionization:
<em><u>1) 0.100 M solution K₂SO₄</u></em>
- K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)
- Mole ratios: 1 mol K₂SO4 : 2 mol K⁺ + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3
- At constant volume, the mole ratios are equal to the concentration ratios (M).
- 1 M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions
This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.
<u>2) 0.0800 M Na₂CO₃</u>
- Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻
- 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions
This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.
<u>3) 0.100 M NaCl </u>
- 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions
This is not equal to 0.300 M ion, so not a correct option.
<u>4) 0.0750 M Na₃PO₄</u>
- 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions
Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.
<u>5) 0.0500 M NaOH </u>
- 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions
Not equal to 0.300 M, so wrong choice.
54 mL Ba(OH)2x(0.101 mol Ba(OH)2/1000 mL) x (2 mol OH-/ 1 mol Ba(OH)2 ) = 0.0109 mol OH-
0.0109 mol OH-x (1mol HCl/ 1 mol OH- ) = 0.0109 mol HCl
0.109 mol HCl/(0.130 mol/L HCl) = 0.0839 L HCl
0.0839 L HCl * 1000mL = 83.9 mL of 0.130 M HCl
Answer:
c
B
B
I Hope u can understand and it's correct answer
True, Crude oil is a type of liquid thereofre can flow through the pourus layer but seeming how impermeable means to not allow fluid to pass it would only make sense that the fluid is trapped between the two.