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2 years ago

Question 6: Iron metal and chlorine gas react to form iron (III) chloride: 2 Fe(s) + 3 Cl2 (g) → 2 FeCl3 (s)

Chemistry

1 answer:

Bas_tet [7]2 years ago

4 0

10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.

What ois Theoretical yield ?

The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.

The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.

As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃

Therefore ,

10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂

As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃

Therefore ,

10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃

Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.

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Answer:

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Explanation:

Given data:

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3 years ago

Which of the following factors could you change in a system and not change its induced EMF?

krok68 [10]

The options attached to the question above are listed below:
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8 0

3 years ago

Read 2 more answers

When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Se

Elodia [21]

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3 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

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5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago