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2 years ago

15

3.04 Quiz: Chemical Reactions Please help me

1 answer:

coldgirl [10]2 years ago

5 0

most of them are chemical because they could not go back to there original form like if you cut paper that is physical if you burn the paper that would be chemical because it would not longer be paper

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If a Ferrari with an initial velocity of 10m/s accelerates at a rate of 5m/s2 for 3 seconds. What will it’s final velocity be?

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Answer:

<em><u>V = 25 m/s</u></em>

Explanation:

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16. What are the reactants in the equation?

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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