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Over [174]
3 years ago
8

How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

700 mL of water

Explanation:

This is the perfect example of dilation calculations. Along with this concept we have a formula c = n ( solute ) / V ( solution ). Let us first solve for n by changing this equation to isolate the solute,

n ( KCL ) = 2.4 mol / L * 500 * 10 ^ - 3 L,

n ( KCL ) = 1.2 moles ( KCL )

Knowing the amount of moles of potassium chloride, we have to now identify how much is present in the target solution,

V = 1 .2 moles /  ( 1.0 moles / L )

V = 1.2 L = 1200 mL

_______________________________________________________

Vadded = 1200 - 500 = 700 mL

<u><em>Hope that helps!</em></u>

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According to question, if Ana has a block made of pure gold and she cuts this block into two equal pieces then the density of each piece is the same as that of the original block because density is an intensive property that does not changes until and unless material is changed.

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\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

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Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

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x = \dfrac{2.6}{100} \times \dfrac{1}{6}

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\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

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K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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