Answer:
C. 2.000 M C6H12O6
Explanation:
Let us obtain the molarity of the solution.
Molar Mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol
Mass of C6H12O6 = 180g
Number of mole = Mass /Molar Mass
Number of mole of C6H12O6 = 180/180 = 1mole
Volume = 500mL = 500/1000 = 0.5L
Molarity = mole /Volume
Molarity = 1/0.5
Molarity = 2M
So the solution will be best labelled as 2M C6H12O6
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.
Answer:
Explanation:
Ketcher 01232019462D 1 1.00000 0.00000 0 5 4 0 0 0 999 V2000 -0.0330 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 2.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 3.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 1.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 3 1 0 0 0 2 4 1 0 0 0 3 5 1 0 0 0 M END