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Andreas93 [3]
2 years ago
12

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

sie Blue followed by the addition of 100 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?
Chemistry
1 answer:
Flauer [41]2 years ago
4 0

Answer:

4.25% is the final concentration of phosphoric acid.

Explanation:

Initial concentration of phosphoric acid = C_1=85\%=0.85

Initial volume of phosphoric acid = V_1=50 mL

Final concentration of phosphoric acid = C_2=?

Final volume of phosphoric acid = V_2=1 L=1000 mL

( 1L = 1000 mL)

C_1V_1=C_2V_2

C_2=\frac{C_1times V_1}{V_2}

=\frac{0.85\times 50 mL}{1000 mL}=0.0425=4.25%

4.25% is the final concentration of phosphoric acid.

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Can someone solve this for me I'm confused.
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Answer:

310.53 g of Cu.

Explanation:

The balanced equation for the reaction is given below:

CuSO₄ + Zn —> ZnSO₄ + Cu

Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:

Molar mass of CuSO₄ = 63.5 + 32 + (16×4)

= 63.5 + 32 + 64

= 159.5 g/mol

Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g

Molar mass of Cu = 63.5 g/mol

Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g

Summary:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.

Thus, 310.53 g of Cu were obtained from the reaction.

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2 years ago
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