2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
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Answer:
Explanation:
48.16%
Well, both abundances have to total 100% so is Ag-107 is 51.84%, then Ag-109 must be 100 – 51.84 = 48.16%.
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Answer:
Explanation:
<u>1) First law of thermodynamic (energy balance)</u>
- Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter
<u>2) Energy change of each substance:</u>
Heat released or absorbed = mass × Specific heat × change in temperature
- density of water: you may take 0.997 g/ ml as an average density for the water.
- mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g
- Specif heat of water: 1 cal / g°C
- Heat released by the hot water:
Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)
- Heat absorbed by the cold water:
Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)
- Heat absorbed by the calorimeter
Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)
<u>4) Balance</u>
49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)
Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K
Ccal = 23.6 cal/ K
- Convert to cal / K to Joule / K
23.6 cal / K × 4.18 J / cal = 98.6 J/K
Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.
Answer:
The answer to your question is 0.269 g of Pb
Explanation:
Data
Lead solution = 0.000013 M
Volume = 100 L
mass = 0.269 g
atomic mass Pb = 207.2 g
Chemical reaction
2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)
Process
1.- Calculate the mass of Pb in solution
Formula
Molarity = 
Solve for number of moles
Number of moles = Volume x Molarity
Substitution
Number of moles = 100 x 0.000013
Number of moles = 0.0013
2.- Calculate the mass of Pb formed.
207.2 g of Pb ----------------- 1 mol
x g ----------------- 0.0013 moles
x = (0.0013 x 207.2) / 1
x = 0.269 g of Pb