Answer:
(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1
(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3
(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7
Explanation:
The rules for electron quantum numbers are:
1. Shell number, 1 ≤ n
2. Sublevel number, 0 ≤ l ≤ n − 1
So,
(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1
(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3
(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals = 2l +1 = 7
Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.
Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.
mass of iron in compound = 34.95 g
molar mass of iron = 55.8 g
mass of oxygen in compound = 15.05 g
molar mass of oxygen = 32 g
number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron
number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1
number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5
the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1
Hence , the required empirical formula of iron compound is FeO.
To learn more about Emiprical formula, refer:
brainly.com/question/1439914
#SPJ4
Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
Answer:
= 15.51 mL
Explanation:
Here's is the reaction:
2HgO(s) ⇒ 2 Hg(s)+O₂(g)
In this reaction 2mol HgO = 1mol O₂
The molecular weight of HgO = 216.59g
so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO
= 0.0138511 molHgO
The amount of Oxygen follows:
0.0138511 molHgOx1/2= 0.00692555 mol O₂
Now, volume of 1 any gas = 22400mL
so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂
= 15.513232mL O₂