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Alex777 [14]
3 years ago
6

Select the reagent that will separate the pairs of ions listed.

Chemistry
1 answer:
daser333 [38]3 years ago
8 0

1 - a, 2-b, 3-d, 4 - e

<u>Explanation:</u>

<u></u>A g^{+} \text {and } A l^{3+} - Separated by 6M HCl

Silver and aluminum ions can be separated using the acid, 6M HCl.

C u^{2+} \text { and } C a^{2+}- Separated by H₂S

Copper and Calcium ions can be separated by the reagent hydrogen sulfide, H₂S.

M g^{2+} \text { and } K^{+} - Separated by (NH₄)₂HPO₄ in NH₃

Magnesium and Potassium ions can be separated by the reagent Ammonium hydrogen phosphate in ammonia.

N i^{2+} \text { and } C o^{2+}- cannot be separated

Nickel and Cobalt ion cannot be separated by the reagents given at all.

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
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Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

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From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

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As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

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Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

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\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
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What is the chemical formula of calcium chloride?
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<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ It would be CaCl2

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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