Answer: The enantiomeric excess ee is 50%
Explanation:
Given that;
compound A = 75%, B = 25%
ee = ?
so
ee = [( A - B) / ( A + B)] × 100
= [( 75 - 25) / ( 75 + 25)] × 100
= [50 / 100] × 100
= 0.5 × 100
= 50%
Therefore enantiomeric excess ee is 50%
Answer:
Because there are already 92 protons in the
uranium nucleus, combining their repulsive power to reject the one new one. However, if the energy of the incoming proton is sufficiently great, then at extremely close distance, the uranium nucleus will attract the proton and draw it into the nucleus. This is called the strong force.
16 S 33
Explanation:
I hope it helped
Answer:
4.5 moles of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
C2H5OH + 3O2 —> 2CO2 + 3H2O
From the balanced equation above,
1 moles of C2H5OH reacted to produce 3 moles of H2O.
Finally, we shall determine the number of mole of H2O produced by the reaction of 1.5 moles of C2H5OH. This can be obtained as follow:
From the balanced equation above,
1 moles of C2H5OH reacted to produce 3 moles of H2O.
Therefore, 1.5 moles of C2H5OH will react to produce = 1.5 × 3/ 1 = 4.5 moles of H2O.
Thus, 4.5 moles of H2O will be produced.
C. The neutron number.
An element is determined based on how many protons has, if the protons number change then you got a new element.
If the number of neuron change you got the same element except it will be a bit heavier or lighter if you are gaining or loosing neutrons. Those variations of an element are called the isotops of the element.
The protons and neutrons toghether form the nucleus of the atom that is heavy and dense as an elephant.
The electrons are light as fleas and stay on the shels of around the nucleus and if they are more then the number of protons they make the atom negatively charged and vice versa.
Answer:
1.0190 x 10⁻⁵ mol
Explanation:
We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).
Molarity = mol/V
V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L
⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L = 1.0190 x 10⁻⁵ mol KIO₃
# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol
