Explanation:
a. Vanillin(4-hydroxy-3-methoxybenzaldehyde):
In its structure hydroxl group will be present on para position of the benzaldehyde ring and methoxy group on meta position.
b. Thymol (2-isopropyl-5-methylphenol):
In its structure isopropyl group will be present on ortho position of the phenol ring and methyl group on meta position.
c. Carvacrol (5-isopropyl-2-methylphenol):
In its structure isopropyl group will be present opposite to methyl group which is present ortho position in a phenol ring.
d. Eugenol (4-allyl-2-methoxyphenol):.
In its structure allyl group will be present on para position of the phenol ring and methoxy group on ortho position.
e. Gallic acid (3,4,5-trihydroxybenzoic acid):
In its structure hydroxyl group will be present on both meta positions and on para position of the benzoic acid ring.
f. Salicyl alcohol (o-hydroxybenzyl alcohol):
In its structure, group is linkedto benzene ring and in respect to that hydroxyl group is present at ortho position of the ring.
Answer:
Tarnish,Striations,Shape,Color,Specific Gravity,Hardness,Ductility.
Explanation:
Ductility: Gold is usually very ductile in nature and will bend on pressure.Fool's Gold will break or resist the pressure.
Hardness: Gold(hardness scale 2.5) will most probably not scratch a copper surface (hardness of 3), but pyrite will very easily be able to do that. Even Gold will be scratched by copper pieces but copper can barely scratch other metals.
Color: Pyrite color is brass type. Gold has a little yellow to golden colour.
Tarnish :God can't be tarnished that easily whereas Pyrite can.
Variables we know:
t = 8 seconds
Vi = 0 m/s
g = -9.81
Δy = ?
Vf = ?
Equation we will be using to solve for Vf: Vf = Vi + gt
Steps to solve:
Vf = (0) + (-9.81)(8)
Vf = -78.48 m/s
Hope this helps!! :)
The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat
Answer:
Average atomic mass = 79.9035 amu.
The given element is bromine.
Explanation:
Given data:
Mass of 1st isotope = 78.9183 amu
Percent abundance of 1st isotope = 50.69%
Mass of 2nd isotope = 80.9163 amu
Percent abundance of 1st isotope = 49.31%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50.69×78.9183 )+(49.31× 80.9163) /100
Average atomic mass = 4000.3686+ 3989.9828 / 100
Average atomic mass = 7990.3514 / 100
Average atomic mass = 79.9035 amu.
The given element is bromine.