Answer:
Q.89
Alkane - CnH(2n+2)
given that 8 H = > 8= 2n+2
therefore n= 3
C3H8 = 12×3 + 8×1= 36 +8 = 44
Despite its appearance, air has a ‘thickness’ so when the sun is high in the sky the light travels through the air on a very much shorter path than when it is low on the horizon.
Imagine that air water and you are below the surface, the light from an overhead sun will be quite sharp and bright, but if lower in the sky it will have to travel through much more water to reach you, so will look less bright and sharp. It ma not seem the same, but the atmosphere is just like very thin water, and a low lying sun will be drastically reduced in strength, so all you will see is a sun with a shift to the red end of the spectrum as all the actinic part will be filtered away by that thicker atmosphere.
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
