STRUCTURE OF BROMOUS ACID: H–O–Br=O
<span>In this structure, all the elements have a formal charge of
zero. The formal charge of each element is calculated below: </span><span>
H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
<span>O: 6 – 1/2(4) – 4 = 0</span></span>
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
There are FOUR electrons in the 3p sub-level of sulfur....
Answer:
The ring of fire
Explanation:
The ring of fire is where they meet or at the fault lines
Answer:
957.7mL
Explanation:
Using the formula below;
CaVa = CbVb
Where;
Ca = concentration of acid (M)
Va = volume of acid (mL)
Cb = concentration of base (M)
Vb = volume of base (mL)
According to the information provided in this question:
Ca = 0.166 M
Cb = 0.013 M
Va = 75mL
Vb = ?
Using CaVa = CbVb
0.166 × 75 = 0.013 × Vb
12.45 = 0.013Vb
Vb =12.45/0.013
Vb = 957.7mL
