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Leviafan [203]
3 years ago
11

If a certain battery supplies 1.0×108 electrons per second to the negative terminal and the battery contains 7.00 moles of elect

rolytic solution (which means the solution contains 7.00 moles of HSO4), what fraction of the solution undergoes a chemical reaction each second
Chemistry
1 answer:
Verdich [7]3 years ago
3 0

Answer:

The answer to the question is

Fraction of solution in reaction = 1.18 × 10⁻¹⁷

Explanation:

The reaction is as follows

H₂SO₄ ⇆ H⁺ + HSO₄⁻

Negative electrode

Pb +  HSO₄⁻ ⇆ PbSO₄ + H⁺ + 2e⁻

That is one mole of H₂SO₄ produces 2 moles of electtrons to the negative electrode

number of electrons in one mole of electrons = Avogadro's number =6.02×10²³

Therefore 7 moles will produce 14 moles of electtrons

Total number of electrons = 14 × 6.02×10²³ = 8.428×10²⁴ electrons

Fraction of electron produced = Fraction of solution in chemical reaction

= 1.0×10⁸÷8.428×10²⁴ = 1.18 × 10⁻¹⁷

Fraction of solution in reaction = 1.18 × 10⁻¹⁷

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Size and Temperature or E & B

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Yield for this reaction?<br> Reaction: N2(g) + 3 H2(g) → 2 NH3 (g)
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Answer:

Yes, yield.

Explanation:

N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation

First, find limiting reactant:  

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The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

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Nonliving things have cells.<br><br> True<br> False
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A balloon (approximate as a sphere) is inflated such that its
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8 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?
kari74 [83]
The heat of reaction (i.e. combustion) of butane (C_{4} H_{10}) when reacted with oxygen (O_{2})  is -2658 kJ/mol butane, and the chemical reaction is given by: 

C_{4} H_{10} + \frac{13}{2} O_{2} ---> 4 CO_{2}  + 5 H_{2}O

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide (CO_{2}) can be determined by multiplying the moles of butane (C_{4} H_{10}) with the mole ratio of (CO_{2}) produced to the (C_{4} H_{10}) reacted, and then with the molar mass of (CO_{2}), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles CO_{2}/ 1 mole C_{4} H_{10}] * 44 grams/mole CO_{2}

Mass of carbon dioxide produced  
    = 99.32 grams CO_{2}

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams. 
                
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