When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃
Answer:
1. KCLO3------>KCL + 3/2O2(g)
2. 122.5g/mol
3. 0.2mol
4. 18.5g
Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
Answer:
the number of neutrons in above isotope = A - Z = 27 - 13 = 14. Note: The molar mass of aluminium, which is average of atomic masses of all isotopes = 26.981538 g/mol, since 13Al27 is the major isotope.
Explanation:
