Answer:
C3 H6 Cl 3
Explanation:
C -24.2%
H - 4.0%
Cl - (100-24.2 - 4.0)=73.8 %
We can take 100g of the substance, then we have
C -24.2 g
H - 4.0 g
Cl - 73.8 g
Find the moles of these elements
C -24.2 g/12.0 g/mol =2.0 mol
H - 4.0 g/1.0 g/mol = 4. 0 mol
Cl - 73.8 g/ 35.5 g/mol = 2.1 mol
Ratio of these elements gives simplest formula of the substance
C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1
CH2Cl
Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol
Real molar mass = 150 g/mol
real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3
So, Real formula should be C3 H6 Cl 3.
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
Answer:
it must be testable I think that's the answer