Answer:
Carbohydrates are polyhydroxy aldehydes, ketones, or compounds that yield polyhydroxy aldehydes or ketones upon hydrolysis. FALSE
Monosaccharides exist as cyclic structures containing functional groups such as hydroxyl groups, acetal (ketal), and hemiacetal (hemiketal). TRUE
A carbohydrate can be classified as a monosaccharide, disaccharide, or polysaccharide TRUE
Disaccharides exist as cyclic structures that only contain hemiacetal (or hemiketal) groups. FALSE
Explanation:
carbohydrates consists of a long cyclic chain of monosaccharides. They are polysaccharides and exist in the cyclic structure. It is only the individual monosaccharides that contain the aldehyde or ketone group as the case may be.
When completely hydrolysed monosaccharides are formed from polysaccharides which are cyclic hemiacetals and acetals containing the hydroxyl functional group.
Carbohydrates containing only one sugar unit is a monosaccharide, if it contains two sugar units it is a diasaccharide, it contains more than two sugar units it is a polysaccharide.
Diasaccharides such as fructose also contain hemiacetal or hemiketal groups and -OH groups also.
Answer:
No, It would be a unsaturated solution
Explanation:
The solubility of a compound gives us information about how a compound may dissolve or not in a determinate solvent.
In this case we have Z, which in 25 °C the solubility of this compound is 40g/100 mL water. This means that if we have 60 g of Z and try to dissolve it in 100 mL of water, only 40 g of Z will solve and the remaining 20 g will be in the water as precipitate or remaining solid.
Now if you just put 40 g of Z in 100 mL water, it will dissolve completely in water, and in this case, we have a saturated solution. A saturated solution is when you dissolve a determinated quantity of a solute in a determinated quantity of solvent, without remaining of solid or excess of solvent.
According to this explanation, we now have 120 g of Z. To make a saturated solution of Z with this quantity, well, let's do math. If 40 g dissolves in 100 mL, then 80 g would be 200 mL and 120 g would have to be 300 mL of water. But in this case, we have 450 mL of water, we have more than 300 mL, an excess of water, so, the 120 g will dissolve but it's dissolved in more than the needed quantity to be a saturated solution, therefore, we have an unsaturated solution of Z (more solvent than the needed).
Hope this helps.
Explanation:
It is known that for 
, ppm present in 1 
are as follows.
1 
= 0.494 ppm
So, 150 
= 
= 0.15
Therefore, calculate the equivalent concentration in ppm as follows.
= 0.074 ppm
Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.
KE = 1/2mv^2
KE = 1/2(4)*10 m/s^2
KE = 2 * 10
KE = 20 joules
Answer: 20 joules
Answer:
A compound can easily be split up into its different elements.
Explanation:
