The empirical formula : C₂Cl₇
The molecular formula : C₁₀Cl₃₅
<h3>Further explanation</h3>
Given
8.81 g Carbon
91.2 g Chlorine
Molar Mass: 1362.5 g/mol
Required
The empirical formula and molecular formula
Solution
Mol ratio :
C = 8.81 g : 12.011 g/mol =0.733
Cl = 91.2 g : 35,453 g/mol = 2..572
Divide by 0.733
C : Cl = 1 : 3.5 = 2 : 7
The empirical formula : C₂Cl₇
(The empirical formula)n = the molecular formula
(C₂Cl₇)n = 1362.5
(2x12.011+7x35.453)n=1362.5
(272.193)n=1362.5
n = 5
The answer is 0.405 M/s
- (1/3) d[O2]/dt = 1/2 d[N2]/dt
- d[O2]/dt = 3/2 d[N2]/dt
- d[O2]/dt = 3/2 × 0.27
- d[O2]/dt = 0.405 mol L^(-1) s^(-1)
Fluorine - Seven electrons of it's own.
Lithium would give up one electron, so there for, fluorine is then left with eight.
No
Water’s density is 1 g/cm3 and AL is 2700kg/m3
<span>Lithium has a property
of high reactivity and to obtain lithium is through electrolysis of its fused
salts. Because lithium is very reactive, it is not found free so electrolysis
is use to split it apart to get it. Moreover,
Lithium is an alkali metal with single valence electron that is easily given up
to form cation, which make it a good conductor of heat and electricity.</span>
<span> </span>