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quester [9]
3 years ago
7

Rick is creating a love potion for Morty. To make the potion, Rick's needs 51 mL of a mixture solution where 40% is carbonated w

ater. After checking around his shop, Rick finds two solutions he could use. The first solution he found is 65% green tea, 15% carbonated water, and 20% whole milk. The second solution is 17% orange juice, 38% lemonade, and 45% carbonated water. How much of the first solution and second solution does Rick need to mix together to create the love potion? Round your final answers to one decimal place. You may solve this problem using any method we have learned in the class.
Chemistry
1 answer:
Deffense [45]3 years ago
8 0

Answer:

The amount of the first solution rick needs to mix together to create the love portion is 8.5 mL

Explanation:

So as to make the love potion, we have;

The percentage of carbonated water in the love portion = 40%

The percentage of green tea in the first solution = 65%

The percentage of carbonated water in the first solution = 15%

The percentage of whole milk in the first solution = 20%

The percentage of orange juice in the second solution = 17%

The percentage of lemonade in the second solution = 38%

The percentage of carbonated water in the second solution = 45%

Let 'x' represent the volume in mL of the first solution added to make the love portion, and let 'y' be the volume in mL of the second solution added to make the love portion, we have;

x + y = 51...(1)

0.15·x + 0.45·y = 0.40×51 = 20.4

0.15·x + 0.45·y = 20.4...(2)

Solving the system of simultaneous equation by making 'y' the subject of each of the equation gives;

For equation (1)

y = 51 - x

For equation (2)

y = 20.4/0.45 - (0.15/0.45)·x = 136 - 3·x

y = 136/3 - (1/3)·x

Equating the two equations of 'y', gives;

51 - x = 136/3 - (1/3)·x

51 - 136/3 = x - (1/3)·x

17/3 = (2/3)·x

(2/3)·x = 17/3

x = (3/2) × (17/3) = 17/2 = 8.5

x = 8.5

y = 51 - x = 42.5

y = 42.5

Therefore, the amount of the first solution rick needs to mix together to create the love portion, x = 8.5 mL

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V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

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M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

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n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
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