B.) Decomposition
hope this helps
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
The bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.
No of moles of calcium bromide is calculated as follows:
moles = 1.642 ÷ 199.89 = 8.215 × 10-³moles
Molarity = 8.215 × 10-³moles ÷ 469.1mL = 1.75 × 10-⁵M
Therefore, the bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
Learn more about concentration at: brainly.com/question/10725862
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<em>Same group element have same</em><em><u> Valence electron</u></em><em> and behave similarly in </em><em><u>Chemistry.</u></em>
<u>Explanation:</u>
For example. First group elements Alkali metals:- H, Li, K, Rb, Cs, Fr
Valance electron will take part in forming a bond with other elements and compound will form. All the above-given elements (H-Fr) have valence electron 1 in outer most 'S' shell. All have electronic configuration S1
Behavior: Since valence electrons are the same so the behavior of all the elements in this group is the same. All are metal (from Li-Fr, except Hydrogen), all are very reactive, does not found in native state in the environment, and all react with water.
