Answer:
it has made the study of chemistry systematic and easy. it acts as an aid to memory
Answer:
The correct answer is option B.
Explanation:
Moles of 
= 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles
Then ,48 moles of NaOH will reacts with:
of
Then ,40 moles of 
will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the is an excessive reagent.
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
Answer: pH = 4.996
Explanation:
No of moles = molarity x volume
:• no of moles of CH3COOH = 0.1M x 0.1L
n(CH3COOH) = 0.1mol
Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.
Therefore, Moles of CH3COOH becomes,
0.1 - 0.03 = 0.07 mol
Subsequently, the moles of CH3COONa increases and becomes,
0.08 + 0.03 = 0.11 mol
Using the Hendersom-Hasselbach equation,
pH = pKa + log [Moles of conjugate÷ moles of Ch3COOH]
From literature, pKa of Ch3COOH is 4.8
Thus,
pH = 4.8 + log [0.11/0.07]
pH = 4.8 + 0.1963
pH = 4.996
