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damaskus [11]
3 years ago
8

At the beginning of the experiment, the aspirin pill contains 325 mg of acetylsalicylic acid and weighs 400 mg. at the end of ex

periment, you obtain 110 mg of pure acetylsalicylic acid. what is the % recovery?
Chemistry
1 answer:
Feliz [49]3 years ago
8 0

That would be (110/ 325)  * 100

= 33.85 %

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How many mL of a stock solution of 2.00 M KNO3 are needed to prepare 100.0 mL of 0.15M KNO3? with work plz
Tatiana [17]
Using the law of dilution :

Mi x Vi =  Mf x Vf

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hope this helps!


3 0
3 years ago
1pt The heaviest concentration of asteroids lies in a belt between
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The answer is A, Mars and Jupiter.
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3 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

6 0
3 years ago
PLEASE<br> What is the H30+ concentration of a solution with a pH of 3?
Alenkinab [10]
PH = -log[H3O+]
Solving for [H3O+] gives
[H3O+] = 10^-pH
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4 0
3 years ago
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