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sergeinik [125]
3 years ago
8

You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions fr

om 12.0 mL of 0.160 M AgNO3 solution? Express your answer with the appropriate units.
Chemistry
1 answer:
padilas [110]3 years ago
4 0

Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

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Answer:

<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

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<u>{ \bf{CH _{3} CH_{2}OH \:  \:  \frac{Ag/O_{2} }{500 \degree C}  >  \:  \:CH _{3} CHO}}</u>

<u>{ \sf{CH _{3} CHO \:  \: is \: ethanal}}</u>

<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

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Which definition best describes global warming?
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What is the solution to the problem expressed to the correct number of significant figures
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I don't know what the problem is, but here are some rues to help you out:

  1. All non-zero figures are significant
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3 years ago
Which statements are true of all scientific endeavors? Check all that apply.
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3 years ago
) How many moles are there in 2.00 x 10^19 molecules of calcium phosphate?
Tamiku [17]

Answer:

\huge 3.322 \times  {10}^{ - 5}  \:  \: moles  \\

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{2.00 \times  {10}^{19} }{6.02 \times  {10}^{23} }  \\  \\  = 3.322 \times  {10}^{ - 5}  \:  \: moles

Hope this helps you

4 0
2 years ago
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