Answer:
C₂H₂O₃
Explanation:
The empirical formula of a compound is derived bu finding the whole ratios of the constituent elements.
In succinic acid, the ratios of carbon to hydrogen to oxygen is calculated as follows:
<u>% mass</u>
Carbon- 40.60
Hydrogen - 5.18
Oxygen - 54.22
<u>RAM</u>
Carbon -12
Oxygen - 15.994
Hydrogen -1.008
<u>No of moles elements in the compound</u>
Carbon = 40.60/12=3.3833
Oxygen = 54.22/15.994= 3.39
Hydrogen= 5.18/1.008 = 5.1389
Mole ratios of the individual elements we divide by the smallest value of the number of moles.
Carbon: Hydrogen : Oxygen
3.3833/3.3833:3.39/3.3833:5.1389/3.3833
=1:1:1.5
We can multiply the value by 2 to get the whole number ratio.
=2:2:3
The empirical formula will be:
C₂H₂O₃
Answer:
pH 8.89
Explanation:
English Translation
If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.
Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹
Assuming all of the salts involved all ionize completely
MgCl₂ ionizes to give Mg²⁺ and Cl⁻
MgCl₂ ⇌ Mg²⁺ + 2Cl⁻
1 mole of MgCl₂ gives 1 moles of Mg²⁺
Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M
Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻
Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²
(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²
[OH⁻]² = (6×10⁻¹¹)
[OH⁻] = √(6×10⁻¹¹)
[OH⁻] = 0.000007746 M
p(OH) = - log [OH⁻] = - log (0.000007746)
pOH = 5.11
pH + pOH = 14
pH = 14 - pOH = 14 - 5.11 = 8.89
Hope this Helps!!!
1 Aluminium is oxidised Al - 3e = Al⁺³
2 Chlorine is reduced Cl⁺⁷ + 8e = Cl⁻¹
3 Nitrogen is oxidised 2N⁻³ - 6e = N₂
Word-Definition
Pure substance- 4
Matter- 2
Atom- 1
Molecule- 3
Element- 5
