Answer:
3
Explanation:
It is based on empirical evidence
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
Here is your answer !!! Last day we have solve this question and the answer is easy to get in mind
Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
