Answer:

Explanation:
Hydrogen bonding:-
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.
Thus, hydrogen must be linked to electronegative atom which is oxygen, fluorine and nitrogen which is in
and thus, it will shown hydrogen bonding.
<u>Answer:</u>
<em>The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock. </em>
<u>Explanation:</u>
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
Answer: Option (b) is the correct answer.
Explanation:
The process in which sediment moves downhill is known as mass movement.
Different types of mass movement are landslides, mud slides, slump, creep etc.
Mud flow contains mass of saturated rock particles of all sizes. Mud flow arises due to sudden flood of water or due to heavy rain in a dry region (semi-arid region). Soil and rocks from a large slope area flow along with the flood water and gets washed to a gulch or canyon.
As a result, debris and water moves down canyon and lay out on the gentle slopes below.
Thus, we can conclude that mud flow is most likely facilitates mass movement after heavy rains in a dry region.
According to the reversible reaction equation:
2Hi(g) ↔ H2(g) + i2(g)
and when Keq is the concentration of the products / the concentration of the reactants.
Keq = [H2][i2]/[Hi]^2
when we have Keq = 1.67 x 10^-2
[H2] = 2.44 x 10^-3
[i2] = 7.18 x 10^-5
so, by substitution:
1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2
∴[Hi] = 0.0033 M