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3 years ago

11

In the reaction below, 22 g of H2S with excess

1 answer:

ella [17]3 years ago

4 0

Answer:

24.1 %

Explanation:

This is the reaction of oxygen between hydrogen sulfide.

The equation is:

2H₂S + O₂ → 2S + 2H₂O

As the oxygen is the excess, limting reagent is the H₂S.

We convert the mass to moleS: 22g / 34.06 g/mol =

0.646 moles

Ratio is 2:2. 2 moles of sulfide can produce 2 moles of sulfur.

Then, 0.646 moles of sulfide will produce 0.646 moles of S.

We convert the moles to mass: 0.646 mol . 32.06 g/mol =

20.71 g

That's the theoretical yield.

% yield = (produced yield / theoretical yield) . 100

% yield = (5 g/ 20.71g) . 100 = 24.1 %

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

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<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

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Here, tin will undergo reduction reaction and will get reduced.

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To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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Putting values in above equation, we get:

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$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

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$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

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3 years ago

How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?

lara [203]

<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

<em>q</em> is heat (in J)
<em>m</em> is mass (in g)
<em>c</em> is specific heat (in J/g °C)
ΔT is change in temperature (in °C or K)

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
Multiply: q = (941.4 J/°C)(159.8 °C)
Multiply: q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

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3 years ago