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.
Technically the are both right and wrong. Solid substances are most likely to sublime is they have... high vapor pressure and weak intermolecular forces.
Answer: 9. decomposition 10. single displacement 11. synthesis 12. double displacement
Explanation: just know I know
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
