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Nikitich [7]
2 years ago
14

2. Which test for iron(II) ions is conclusive ​

Chemistry
2 answers:
krok68 [10]2 years ago
5 0

Answer:

please brainlist answer

Explanation:

The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.

melomori [17]2 years ago
4 0

Answer:

the addition of k3 Fe(CN)6 to a solution causes the formation of deep blue precipetate which indicates iron(ll) ions

are present

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What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?
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Answer:

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A 2.0 molal sugar solution has approximately the same freezing point as 1.0 molal solution of 1) CaCl2 2) CH3COOH 3) NaCl 4) C2H
sertanlavr [38]

Answer:

3) NaCl.

Explanation:

<em>∵ ΔTf = iKf.m</em>

where, <em>i</em> is the van 't Hoff factor.

<em>Kf </em>is the molal depression freezing constant.

<em>m</em> is the molality of the solute.

<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>

<em></em>

  • For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<em>So, for sugar: i = 1.</em>

<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>

<em></em>

  • For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.

<em>So, i for NaCl = 2.</em>

<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>

<em></em>

<em>So, the right choice is: 3) NaCl.</em>

<em></em>

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