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ioda
3 years ago
6

A chef plans to mix 100% vinegar with italian dressing. the italian dressing contains 13% vinegar. the chef wants to make 150 mi

lliliters of a mixture that contains 42% vinegar. how much vinegar and how much italian dressing should she use?
Chemistry
1 answer:
charle [14.2K]3 years ago
6 0
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
y = 100

Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.

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<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

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\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

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The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

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\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

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