Answer:
N₂ + 3H₂O → 2NH₃ + 3NO
Explanation:
The reaction expression is given as:
N₂ + H₂O → NH₃ + NO
We are to balance this reaction expression.
To do this, assign a,b,c and d as the coefficients that will balance the expression:
aN₂ + bH₂O → cNH₃ + dNO
Conserving N: 2a = c
H: 2b = 3c
O: b = d
let a = 1, c = 2, b = 3 and d = 3
So;
N₂ + 3H₂O → 2NH₃ + 3NO
Firstly need to determine the empirical formula of the hydrocarbon. Empirical formula is the simplest whole number ratio of components of the compound. Molecular formula is the actual composition of the components in the compound.
percentage of C - 82.66%
percentage of H - (100-82.66) = 17.34 %
in 100 g of compound ;
mass of C - 82.66 g
mass of H - 17.34 g
C H
mass in 100 g 82.66 g 17.34 g
molar mass 12 g/mol 1 g/mol
number of moles 6.88 mol 17.34 mol
(mass/molar mass)
divide the number of moles by least number of moles (6.88 mol)
6.88 mol/6.88 17.34/6.88
1 2.52
multiply these by 2 to get a whole number
C - 1x 2 = 2
H - 2.52 x 2 = 5.04
round off to nearest whole number
C - 2
H - 5
ratio of C to H is 2:5
empirical formula - C₂H₅
empirical formula mass = 12 g/mol x 2 + 5 * 1 g/mol = 29 g
next have to find how many empirical units are there in the molecular unit
molecular unit mass = 58.12 g
empirical unit = 29 g
then number of empirical units = 58.12 / 29 = 2
rounded off , number of empirical units = 2
(C₂H₅) * 2 units
molecular formula = C₄H₁₀
Explanation:
Answer:
diffusion
Explanation:
Diffusion can be defined as the net movement of particles from a high concentration to a low concentration. This sounds like the most plausible word in this case.
The balanced equation for the above reaction is as follows;
3NaOH + H₃PO₄ ---> Na₃PO₄ + 3H₂O
stoichiometry of NaOH to H₃PO₄ is 3:1
the number of NaOH moles in the solution - molarity x volume
number of NaOH moles - 0.3 mol/L x 0.030 L = 0.009 mol
for complete neutralisation
3 mol of NaOH requires 1 mol of H₃PO₄
therefore 0.009 mol of NaOH requires - 1/3 x 0.009 = 0.003 mol of H₃PO₄
therefore 0.003 mol of H₃PO₄ are needed to reach the equivalence point
By using ICE table:
CH3NH3+ + H2O → CH3NH4 2+ + OH-
initial 0.175 0 0
change -X +X +X
Equ (0.175-X) X X
when: Ka = Kw / Kb
= (1 x 10^-14) / (4.4 x 10^-4) = 2.3 x 10^-11
when Ka = [CH3NH42+][OH-] / [CH3NH3+]
by substitution:
2.3 x 10^-11 = X^2 / (0.175 - X ) by solving for X
∴ X = 2 x 10^-6
∴[OH-] = 2 x 10^-6
∴POH = -㏒[OH-]
= -㏒(2 x 10^-6)
= 5.7
when PH + POH = 14
∴PH = 14 - 5.7 = 8.3