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a_sh-v [17]
3 years ago
13

Find the mass of each of these substances? 2.40 mol NaOH

Chemistry
1 answer:
damaskus [11]3 years ago
4 0
Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40

55.2 + 38.4 + 2.4 = 96

2.40 mol of NaOH = 96 amu

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This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH  + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
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Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol 
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
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When an aqueous solution of magnesium nitrate is mixed with an aqueous solution of potassium carbonate, ____________.?
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A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
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Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

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<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

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<u>Step 3:</u> Calculate total mass

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<u>Step 4</u>: Calculate moles of AgBr

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1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

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