Answer:
ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C
given the specific heat of iron as 0.108 cal/g·°C
heat=(100.0 g)(0.108 cal /g· °C )(125°C) =
100x 0.108x125= 1350 cal
Answer:
The boiling point elevation is 3.53 °C
Explanation:
∆Tb = Kb × m
∆Tb is the boiling point elevation of the solution
Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m
m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram
Moles of solute = mass/MW =
mass = 92.7 mg = 92.7/1000 = 0.0927 g
MW = 132 g/mol
Moles of solute = 0.0927/132 = 7.02×10^-4 mol
Mass of solvent = 1 g = 1/1000 = 0.001 kg
m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg
∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)
Answer:
Glycogen. Cellulose. Amylose. Cellulose. Amylopetin and Glycogen. Amylopetin and Cellulose.
Explanation:
Glycogen is the form that glucose is stored in human body.
Cellulose is the structural part of plant cell walls and human cannot digest it.
Amylose is the polysaccharide linked mainly by the the bonds of 
1,4 glycosidic.
Cellulose is an unbranched polysaccharide linked mainly by the bonds of 
1,4 glycosidic.
Amylopetin and Glycogen are branched polysaccharides linked by the bonds of 1,4 glycosidic and 1,6 glycosidic.
Amylopetin and Cellulose are mainly stored in plants.
I think the answer will be b cuh
