Answer:
I can't say that it is definitely write.
HHH
H-C-C-C
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
Answer:
Empirical formula = K₂SO₄
The name of the compound is Potassium tetraoxosulfate (vi)
Explanation:
Percentage composition of the elements:
Potassium = 44.82%
Sulfur = 18.39%
Oxygen = 36.79%
Mole ratio of the elements:
Potassium = 44.82/39 =1.15
Sulfur = 18.39/32 = 0.57
Oxygen = 36.79/16 = 2.29
Dividing by the smallest ratio in order to obtain the simplest ratio
Potassium = 1.15/0.57 =2
Sulfur = 0.57/0.57 = 1
Oxygen = 2.29/0.57 = 4
Therefore, the mole ratio of Potassium : Sulfur : Oxygen = 2:1:4
Empirical formula = K₂SO₄
The name of the compound is Potassium tetraoxosulfate (vi)
I think it is D but i'm not sure
hope I helped :)
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in 
of ethane gas
And, as we know that
1 mole of ethane molecule contains 
molecules of ethane
2.869 moles of ethane molecule contains 
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
