How many liters of water should be added to 20 liters of 40% solution of acid to

1 answer:

8 0

The amount of liters of water that should be added to 20 liters of 40% solution of acid to obtain 30% solution is: 6.7 Liters.

<h3>Liters of water needed</h3>

Acid = 0.4× 20

Acid = 8 liters

Water = 0.6× 20

Water = 12 liters

Acid/(Acid+Water)

A / (A + W) = 0.30

8 / (8 + W) = 0.30

2.4 + 0.3W = 8

0.3W = 8 - 2.4

0.3W= 5.6

Divide both side by 0.3W

W=5.6/0.3

W=18.7 Liters

18.7 - 12 = 6.7 Liters (Amount of water that must be added)

Therefore the amount of liters of water that should be added to 20 liters of 40% solution of acid to obtain 30% solution is: 6.7 Liters.

Learn more about Liters of water needed here: brainly.com/question/23918882

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If the volume of a sphere is equal to the volume of a cube, what is the ratio of the edge of the cube to the radius of the spher

Pachacha [2.7K]

Answer: The answer is (C) 1.61.

Step-by-step explanation: Given that the volume of a sphere is equal to the volume of a cube. We are to find the ratio of the edge of the cube to the radius of the sphere.

Let, 'r' be the radius of the sphere and 'e' be the edge of the cube. Then, the volume of the sphere is given by

$V_s=\dfrac{4}{3}\pi r^3,$

and the volume of the cube is

$V_c=e^3.$

According to the question, we have

$V_s=V_e\\\\\Rightarrow \dfrac{4}{3}\pi r^3=e^3\\\\\\\Rightarrow \dfrac{4}{3}\times\dfrac{22}{7}r^3=e^3\\\\\\\Rightarrow 88r^3=21e^3\\\\\Rightarrow 4.44r=2.75e\\\\\Rightarrow \dfrac{e}{r}=\dfrac{4.44}{2.75}\\\\\Rightarrow e:r=1.61.$