The molecular formula :
C₆H₁₄O₃PF
<h3>Further explanation</h3>
Given
39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine.
Required
The molecular formula
Solution
mol ratio :
C = 39.1 : 12 = 3.258
H = 7.67 : 1 = 7.67
O = 26.11 : 16 = 1.632
P = 16.82 : 31 = 0.543
F = 10.3 : 19 = 0.542
Divide by 0.542
C = 6
H : 14
O = 3
P = 1
F = 1
The empirical formula :
C₆H₁₄O₃PF
(The empirical formula)n = the molecular formula
(C₆H₁₄O₃PF)=184.1
(6.12+14.1+3.16+31+19)n=184.1
(184)n=184.1
n = 1
Answer:
bent
Explanation:
The molecular formula of sulfur dioxide is written as SO₂
The molecular geometry of sulfur dioxide can be determined using the Lewis structure.
The Lewis structure shows the distribution of electrons around the atoms of a given compound such as sulfur dioxide (SO₂).
In this compound, sulfur is the central atom with 6 valence electrons.
The sulfur is bonded covalently with two oxygen atoms, each with 6 valence electrons. Oxygen contributes 2 lone pairs while sulfur which is the central atom contributes 1 lone pair of electrons in the bond.
The bond angle between the two oxygen atoms and the central sulfur atom is approximately 120⁰, as a result of the bent shape of the molecular structure.
Answer:
40 C atoms
Explanation:
Step 1: Write organic compound
10C₄H₁₀
We see here that in 1 mol of C₄H₁₀, we would have 4 atoms of C.
We are given that we have 10 moles of C₄H₁₀.
Therefore, we have 4 · 10 = 40 atoms C
The answer is (c) lessen the environmental impact of acid rain. This is because acid rain is mainly due to the <span>sulfur dioxide (and nitrogen oxides) emissions from smokestacks (and motor vehicles). Controlling the sulfur dioxide emissions would mean an attempt to lessen the environmental impact of acid rain.</span>
<span>The value of enthalpy of a reaction can be a useful tool in determining the type of reaction it is. If the enthalpy is negative the reaction is exothermic and if the enthalpy is positive the reaction is endothermic. Hope this answers the question. Have a nice day.</span>
