Answer:
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Explanation:
Answer:
Mass = 18.9 g
Explanation:
Given data:
Mass of Al₂O₃ formed = ?
Mass of Al = 10.0 g
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 27 g/mol
Number of moles = 0.37 mol
Now we will compare the moles of Al and Al₂O₃.
Al : Al₂O₃
4 : 2
0.37 : 2/4×0.37 = 0.185 mol
Mass of Al₂O₃:
Mass = number of moles × molar mass
Mass = 0.185 mol × 101.9 g/mol
Mass = 18.9 g
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
Answer:
The death is lower
Explanation:
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