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ra1l [238]
3 years ago
10

What is the freezing point, in °C, of a solution made with 1.31 mol of CHCl₃ in 530.0 g of CCl₄?

Chemistry
1 answer:
lutik1710 [3]3 years ago
7 0

Answer: 73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.A

Explanation: brainliest?

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A chef plans to mix 100% vinegar with italian dressing. the italian dressing contains 13% vinegar. the chef wants to make 150 mi
charle [14.2K]
To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns. 

For the first equation, we do a mass balance:

mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar

Assuming they have the same densities, then we can write this equation in terms of volume.

V(100%) + V(13%) = V(42%)
   we let x = V(100%)
             y = V(13%)

x + y = 150

For the second equation, we do a component balance:

1.00x + .13y = 150(.42)
x + .13y = 63

The two equations are
x + y = 150
x + .13y = 63

Solving for x and y,
x = 50
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Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.

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