What distance does electromagnetic radiation travel in 33 fs?

Given:

Time taken by the electromagnetic radiation, t = 33 fs

To determine:

The distance (d) travelled

Explanation:

Time taken = 33 fs (femtosecond)

1 fs = 10⁻¹⁵ s

Thus t = 33 *10⁻¹⁵ s

Speed of electromagnetic radiation = 3*10⁸ m/s

Distance = Speed * Time = 3*10⁸ ms-1 * 33*10⁻¹⁵ s = 99*10⁻⁷ m

Ans: distance traveled = 9.9*10⁻⁶ m

distance traveled = 9.9*10⁻⁶ m

Explanation:

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.