Answer:
A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.
Explanation:
Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.
An example of a polysaccharide is starch, or glycogen.
Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.
Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.
Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.
While terrestrial biomes are shaped by air temperature and precipitation, aquatic systems are characterized by factors such as water salinity, depth, and whether the water is moving or standing. If that's what you mean?
calcium,phosphorus,potassium,and sulfer
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
An ester , propyl methanoate ( HCOOC₃H₇) when reacts with sodium hydroxide( NaOH) forms sodium methanoate (HCOONa) as the main product and propanol (C₃H₇OH).
The reaction is as follows:
HCOOC₃H₇+NaOH ⇒HCOONa + C₃H₇OH
So when propyl methanoate is hydrolyzed in water and in NaOH then sodium methanoate (HCOONa) as the main product and propanol (C₃H₇OH) forms