Answer:
then why'd you ask a question
Explanation:
Q1)
firstly we need to determine the empirical formula of the compound. empirical formula is the simplest ratio of components in the compound.
percentages of the elements have been given, so lets assume we are calculating for a compound of 100g
C H O
mass 63.13 g 8.830 g 28.03 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.13/12 8.830/1 28.03/16
5.26 8.830 1.75
divide by the smallest number of moles
5.26/1.75 8.830/1.75 1.75/1.75
= 3.01 = 5.04 =1
rounded off to the nearest whole numbers
C - 3
H - 5
O - 1
therefore empirical formula = C₃H₅O
Q2)
we have to next determine the molecular formula of the compound
molecular formula gives the actual composition of elements in the compound.
since we know the empirical formula and molecular mass, we can find how many empirical units are in the molecular formula.
mass of empirical unit = Cx3 + Hx5 + Ox1
= 12 g/mol x 3 + 1g/mol x 5 + 16 g/mol x 1
= 36 + 5 + 16 = 57 g/mol
the molecular mass = 228 g/mol
then number of empirical units in the molecular formula = 228 / 57 = 4
therefore there are 4 empirical units
then the molecular formula = 4 x empirical formula =4 (C₃H₅O)
molecular formula = C₁₂H₂₀O₄
It's simple.
the number of protons in an atom never changes. And also, the no of protons is its atomic number.if you look up the periodic table, you will find a number on the top - left corner of each of the boxes. This is the atomic number.No 2 same elements on the periodic table have the same atomic number. So, all you have to do is search for a box that has the number 13 on it.
and by the way ... the element which you're searching for is Aluminium
Hello!
The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29](https://tex.z-dn.net/?f=Ka%3D%20%5Cfrac%7B%5BC_6H_5COO%5E%7B-%7D%5D%2A%5BH_3O%5E%7B%2B%7D%5D%20%7D%7B%5BC_6H_5COOH%5D%7D%3D%5Cfrac%7BX%2AX%20%7D%7B0%2C40%20-X%7D%20%20%28assume%3A%200%2C40-X%5Capprox0%2C40%29%5C%5C%20%20%5C%5C%20X%3D%20%5Csqrt%7B0%2C40%2A6%2C50%2A10%5E%7B-5%7D%20%7D%3D0.00510M%20%5C%5C%20%20%5C%5C%20pH%3D-log%28%5BH_3O%5E%7B%2B%7D%5D%3D2%2C29%20%20%20)
So, the pH of this Benzoic Acid solution is
2,29Have a nice day!
