Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer:
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Explanation:
Answer:
92%
Explanation:
I believe. Hope this was helpful.
173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it
When drawing lewis dot diagram structure we consider the number of valence electrons of the atom.Se has six valence electrons since it is in group six thus it Lewis dot diagram is as follows
..
: Se:
Hydrogen is in group one hence has one valence electron.The lewis dot diagram for 2H is therefore
H:H