Answer:
See explanation
Explanation:
A. Constitutional or structural isomers have the same molecular formula but different structural formulas.
B. Conformational isomers are compounds having the same atom to atom connectivity but differ by rotation about one or more single bonds.
C. Stereo isomers are compounds having the same molecular mass and atom to atom connectivity but different arrangement of atoms and groups in space.
I. Enantiomers are stereo isomers (optical isomers particularly) that are non-superimposable mirror images of each other.
II. Diasteromers are optical isomers that are not mirror images of each other.
Both diasteromers and enantiomers are types of optical isomers which in turn is one of the types of stereo isomers.
Stereo isomers differ from conformational isomers in that the arrangement of atoms in stereo isomers is permanent while conformational isomers results from free rotations in molecules about single bonds.
Consider you have a mixture of amino acids(contains all set of amino acids such as polar, non polar). Other than TLC, how are you supposed to separate a single amino acid from the mixture without loss of amino acid quantitatively.
A mixture of oil and water is:
Heterogeneous
immiscible
<h3>What type of mixture of oil and water is?</h3>
The water molecules attract each other, and the oil molecules stick together. That causes oil and water to form two separate layers. Water molecules pack closer together, so they sink to the bottom, leaving oil sitting on top of the water. The mixture can be called as Emulsion. It is an immiscible liquid and the mixture is heterogeneous as oil floats on top of water surface but do not mix with it.
Thus, correct options are Heterogeneous and immiscible.
Find more information about Mixture here:
brainly.com/question/498370
Missing in your question :
Ksp of(CaCO3)= 4.5 x 10 -9
Ka1 for (H2CO3) = 4.7 x 10^-7
Ka2 for (H2CO3) = 5.6 x 10 ^-11
1) equation 1 for Ksp = 4.5 x 10^-9
CaCO3(s)→ Ca +2(aq) + CO3-2(aq)
2) equation 2 for Ka1 = 4.7 x 10^-7
H2CO3 + H2O → HCO3- + H3O+
3) equation 3 for Ka2 = 5.6 x 10^-11
HCO3-(aq) + H2O(l) → CO3-2 (aq) + H3O+(aq)
so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3-(aq)
note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :
CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l) Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq) + CO3-2(aq) Ksp = 4.5 x 10^-9
∴ the overall equation will be as we have mentioned before:
when H3O+ = H+
CaCO3(s) + H+(aq) ↔ Ca2+ (aq) + HCO3-(aq) K= 80.55
from the overall equation:
∴K = [Ca2+][HCO3-] / [H+]
when we have [Ca2+] = [HCO3-] so we can assume both = X
∴K = X^2 / [H+]
when we have the PH = 5.6 so we can get [H+]
PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6
so, by substitution on K expression:
∴ 80.55 = X^2 / (2.5 x10^-6)
∴X = 0.0142
∴[Ca2+] = X = 0.0142
The original or accepted value for the percent by mass of water in a hydrate = 36%
Percen by mass of water in the hydrate determined by the student
in the laboratory = 37.8%
So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%
= 1.8%
So the percentage of error made by the student = (1.8/36) * 100 percent
= (18/360) * 100 percent
= ( 1/20) * 100 percent
= 5 percent
So the student makes an error of 5%. Option "1" is the correct option.
