How many grams of ag can be formed from 5.50 grams of ag2o in the equation: 2ag2o (s) → 4ag (s) o2 (g)?
1 answer:
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O== 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
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Answer:
Final temperature: 659.8ºC
Expansion work: 3*75=225 kJ
Internal energy change: 275 kJ
Explanation:
First, considering both initial and final states, write the energy balance:
Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:
The pressure is constant, so: (There is a multiplication by 100 due to the conversion of bar to kPa)
So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):
On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:
Subtracting the first from the second:
Isolating :
Assuming that it is water steam, n=0.1666 kmol
ºC