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son4ous [18]
3 years ago
11

If the compound potassium chloride is written in words, how do you know to write the formula with one potassium and one chloride

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
6 0

Because there is only one stable ionic compound made up of potassium and chlorine, and that is KCl. So calling is "mono chloride" or similar would be redundant assuming you understand basic chemistry (i.e. knowing oxidation numbers of K is +1 and Cl is -1). When compounds can exist in multiple forms in nature like CO and CO2 you will preferably indicate it through the nomenclature, calling one a monoxide and the other a dioxide.

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Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
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Answer:1) Volume of AgNO_3 required is 55.98 mL.

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Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

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Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

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Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

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