The empirical formula of the compound is C4H1.
An empirical formula represents the whole number ratio of various atoms present in a compound.
If the mass percent of various elements present in a compound is known, its empirical formula can be determined
We are given that a compound contains 4 mole C and 1 mole H.
Hence, the ratio of C: H is 4:1, which is already in the form of the simplest whole number.
Therefore, the empirical formula of the given compound is C4H1 and its molecular mass is 12 x 4 + 1 = 49.
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Answer:
2s² 2p⁵
There are seven valance electrons in fluorine.
Explanation:
The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other.
Properties of fluorine:
1. it is yellow in color.
2. it is flammable gas.
3. it is highly corrosive.
4. fluorine has pungent smell.
5. its reactions with all other elements are very vigorous except neon, oxygen, krypton and helium.
Electronic configuration of fluorine:
F₉ = 1s² 2s² 2p⁵
Valance electrons in fluorine are 2s² 2p⁵.
Valance Orbital configuration:
2s² 2p⁵
There are seven valance electrons in fluorine.