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:
MnO4- + 5 VO +2 + 11 H2O = 5 V(OH)4+ + Mn+2 + 2 H+
Explanation:
Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
<em><u>Answer</u></em><em><u> </u></em>: HBr polar
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