Answer: Amylose is a form of starch which has only α-1,4-links bonds glucose units.
Explanation:
Amylose is a polysaccharide made up of α(1-4) bound glucose molecules. The carbon atoms on glucose are numbered, starting at the aldehyde (C=O) carbon, so, in amylose, the 1-carbon on one glucose molecule is linked to the 4-carbon on the next glucose molecule.
Democritus was the one who had theorized that atoms make up everything and they are indivisible.
Dalton was the creator of the first actual atomic theory, most of his research was on gasses and meteorology.
Thompson was the original man who put together the plum pudding model in which Rutherford later proved wrong during his career.
Rutherford had discovered the nucleus within an atom. He had put together gold foil experiment.
Bohr had developed the idea of neutrons and electrons surrounding the nucleus. He was also the creator of the planetary model we now use to calculate electrons with.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
the answer is b Li + Cl2 .....
0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.
Explanation:
Data given:
volume of the nitrogen gas = 2 litres
Standard temperature = 273 K
Standard pressure = 1 atm
number of moles =?
R (gas constant) = 0.08201 L atm/mole K
Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law
PV = nRT
rearranging the equation to calculate number of moles:
PV = nRT
n = 
putting the values in the equation:
n = 
n = 0.091 moles
0.091 moles of nitrogen gas is contained in a container at STP.
