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saul85 [17]
2 years ago
15

During a laboratory experiment, 9.68 g of iron reacted with excess sulfur to form 13.8 grams of iron II sulfide. Calculate the p

ercent yield of this experiment. What would be the percentage yield if the Theoretical yield was 15.2 grams?
Chemistry
1 answer:
Daniel [21]2 years ago
5 0

Answer:

Percent yield = 90.8%

Explanation:

The reaction of Fe with S to produce FeS is:

Fe + S → FeS

<em>Where the moles of Fe added in excess of S are the moles of FeS</em>

<em />

Now, percent yield is defined as 100 times the ratio between actual yield (13.8g of FeS) and theoretical yield (15.2g FeS):

Percent yield = 13.8g / 15.2g * 100

<h3>Percent yield = 90.8%</h3>
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MissTica
Hello there.

Question: <span>In order to complete a convection current, the rising material must eventually ____ Earth.

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3 years ago
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What is the hybridization and shape for an XeF6 2+ molecule?
Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

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Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

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BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

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2 years ago
Name:
romanna [79]

Answer:

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Explanation:

Hope it helps.

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2 years ago
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
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Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

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8 0
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What is required for electrons to move from the stable energy level to an excited state?
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answer

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