i performed this experiment and it comes out around 60 (radians*100cm3/gm*dm) where length of polarimeter tube was 2 dm and concentration was varied from 40 gm/100cm3 to 20gm/100cm3
when sucrose is hydrolysed it turns into glucose and fructose. fructose is laevo rotatory whereas glucose is dextrorotatory sucrose is also dextrorotatory but amout of laevo rotation of fructose is able to overcome the dextro rotation of glucose. hence the product is laevo rotatory an d reactant dextrorotatory hence sucrose is an invert sugar.
If 180g glucose is present in one litre of solution then boiling point is 100.52 Celsius.
Glucose is a monosaccharide that serves mainly as a food molecule.
because he is the Almighty, he can move mountains ,that is why!
Fat or more widely known as lipids is not soluble to water glucose is soluble in water.you need to be more specific as to what you think they migth be soluble in.
10 mg of glucose
.33
75 degrees Celsius
The formation of glucose slows down after the temperature reaches 30 degrees Celsius.
4.2(novanet)
A rise in temperature will increase the rate of photosynthesis, but over 40 degrees will rapidly decrease the rate of photosynthesis. I'm not entirely certain, but I would guess about 15-25 degrees is best.
maltose, its products are glucose, the organ it is used in is duodenum, its optimal pH is 6.1-6.8, and its optimal temperature is 35-40 degrees Celsius.
when sucrose is hydrolysed it turns into glucose and fructose. fructose is laevo rotatory whereas glucose is dextrorotatory sucrose is also dextrorotatory but amout of laevo rotation of fructose is able to overcome the dextro rotation of glucose. hence the product is laevo rotatory an d reactant dextrorotatory hence sucrose is an invert sugar.
If 180g glucose is present in one litre of solution then boiling point is 100.52 Celsius.
Sugar (or glucose)
250 degrees.
50.5g of glucose glucose is 180.18 g/mol 50.5/180.18 = 0.280 mol glucose 0.280 mol glucose/0.475 kg H2O = 0.589 m which is molality deltaTsubF = (-1.86 degress C/m)(0.589) = -1.10 degrees C -1.10 degrees C + 0.00 degrees C = -1.10 degrees C as your freezing point seeing that 0.00 degrees is the standard freezing point of water deltaTsubB = (0.512 degrees C/m)(0.589) = 0.301568 it asks for 6 sig figs so 0.301568 degrees C + 100.000 degrees C (boiling point H2O) = 100.302 degrees C Mastering Chemistry sucks sometimes....