Answer:
(1) -12 Kcal/mol
Explanation:
Our answer options for this question are:
(1) -12 Kcal/mol
(2) -13 Kcal/mol
(3) -15 Kcal/mol
(4) -16 Kcal/mol
With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, <u>two bonds are broken</u>, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are <u>formed</u>.
If we want to calculate the enthalpy value, we can use the equation:
<u>ΔH=ΔHbonds broken-ΔHbonds formed</u>
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
<u>ΔHbonds broken</u>
<u />
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
<u>ΔHbonds formed</u>
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
<u>ΔH of reaction</u>
<u />
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it helps!
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D. I or Iodine
This is because they are both in the same periodic family (the halogens) and thus the number of valence electrons are the same
Answer:

Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula 
Answer:
Group 18th
Explanation:
The group 18th of periodic table which is composed of Noble gases mainly was missing from the Mendeleev's periodic table. The reason for this is that Noble gases are least reactive, therefore, it was difficult to predict such elements which do not react with other elements.