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Step2247 [10]
3 years ago
9

How many liters of a 6.0M solution Fe(SCN)2 are needed to prepare 2.0L of a 0.75M solution?

Chemistry
2 answers:
scoray [572]3 years ago
8 0

Answer:

We need 0.25 L of the 6.0 M solution

Explanation:

Step 1: Data given

Molarity of the Fe(SCN)2 solution = 6.0 M

Volume of the 0.75 M solution = 2.0 L

Step 2: Calculate volume

C1*V1 = C2*V2

⇒ with C1 = the initial concentration = 6.0 M

⇒ with V1 = the initial volume = TO BE DETERMINED

⇒ with C2 = the new concentration = 0.75 M

⇒ with V2 = the new volume = 2.0 L

6.0M * V1 = 0.75M * 2.0L

V1 = (0.75 M * 2.0L) / 6.0 M

V1 = 0.25 L

We need 0.25 L of the 6.0 M solution

lubasha [3.4K]3 years ago
4 0

Answer:

0.25L

Explanation:

Using the dilution formula

C1V1=C2V2

C1=6M

V1?

C2=0.75M

V2=2.0L

V1= C2V2/C1

V1=0.75*2.0/6

V1=0.25L

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Fudgin [204]
The chemical reaction would be expressed as follows:

HBr + LiOH = LiBr + H2O

We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:

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A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

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Hence, if solubility of Pb(IO_{3})_{2} is S (M) then-

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Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

So, [IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M

So, K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}

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Explanation:

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Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.

For example for the given equilibrium reaction;

2H_2O(g)\leftrightharpoons 2H_2(g)+O_2(g)

K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

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