Answer:
138.96kJ is the maximum electrical work
Explanation:
The maximum electrical work that can be obtained from a cell is obtained from the equation:
W = -nFE
<em>Where W is work in Joules,</em>
<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>
Zn(s) → Zn²⁺(aq) + 2e⁻
F is faraday constant = 96500Coulombs/mol
E is cell potential = 0.72V
Replacing:
W = -2mol*96500Coulombs/mol*0.72V
W = - 138960J =
<h3>138.96kJ is the maximum electrical work</h3>
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Answer:
5
Explanation:
They have 5 in common but different x
Answer:
90g of H2O
Explanation:
2H2 + O2 —> 2H2O
First, we calculate the molar masses of H2 And H20.
Molar Mass of H2 = 2g/mol
Mass conc of H2 from the balanced equation = 2 x 2 = 4g
Molar Mass of H2O = 2 + 16 = 18g/mol
Mass conc of H2O from the balanced equation = 2x18 = 36g
From the equation,
4g of H2 produced 36g of H2O
Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:
Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g
It would be 7 because the acid and base cancel out each other
