Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
Answer:
NH₃/NH₄Cl
Explanation:
We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.
![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.
Let's consider the following buffers and their pKa.
- CH₃COONa/CH3COOH (pKa = 4.74)
The optimum buffer is NH₃/NH₄Cl.
Answer:
volume in L = 0.25 L
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 2.43 g
Volume of KI = ?
Solution:
Balanced chemical equation:
2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 2.43 g/ 187.56 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of Cu(NO₃)₂ with KI.
Cu(NO₃)₂ : KI
2 : 4
0.013 : 4 × 0.013=0.052 mol
Volume of KI:
<em>Molarity = moles of solute / volume in L</em>
volume in L = moles of solute /Molarity
volume in L = 0.052 mol / 0.209 mol/L
volume in L = 0.25 L
Answer:
ΔG°rxn = -72.9 kJ
Explanation:
Let's consider the following reaction.
HCN(g) + 2 H₂(g) → CH₃NH₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:
ΔG°rxn = ΔH° - T.ΔS°
where,
ΔH° is the standard enthalpy of the reaction
T is the absolute temperature
ΔS° is the standard entropy of the reaction
ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)
ΔG°rxn = -72.9 kJ
The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below
write the equation for the dissociation of Ba(OH)2
Ba(OH)2 = Ba^2+ +2OH^-
calculate the OH- concentration
by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M
by use of the formula ( H3O+)(OH-) = 1 x10 ^-14
by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-
substitute for OH-
H3O+ = (1 x10^-14 )/0.1
= 1 x10^-3 M
