According to this equation:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
so K1 = [Ag+][Cl-]
when [Ag+] = [Cl-] we can assume both = X
and when we have X the solubility = 1.33 x 10^-5 mol / L
by substitution:
∴ K1 = X^2
= (1.33 x 10^-5)^2
= 1.77 x 10^-10
by using vant's Hoff equation:
ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)
when ΔH = 65700 J / mol
R = 8.314
T1 = 25+273 = 298 K
T2 = 47.7 +273 =320.7
by substitution:
∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)
by solving for K2
∴K2 = 2.7 x 10^-11
and when K2 = X^2
∴ the solubility X = √(2.7 x 10^-11)
= 5.2 x 10^-6 mol/L
Chemistry affects all aspects of life and most natural events because all living and nonliving things are made of matter.
Answer:
i think it's eutrophication
Answer:
85.6 g
Explanation:
Step 1: Write the balanced combustion equation
C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O
Step 2: Calculate the moles corresponding to 140 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
140 g × 1 mol/18.02 g = 7.77 mol
Step 3: Calculate the moles of C₃H₈ needed to produce 7.77 moles of H₂O
The molar ratio of C₃H₈ to H₂O is 1:4. The moles of C₃H₈ needed are 1/4 × 7.77 mol = 1.94 mol.
Step 4: Calculate the mass corresponding to 1.94 moles of C₃H₈
The molar mass of C₃H₈ is 44.10 g/mol.
1.94 mol × 44.10 g/mol = 85.6 g
For any given shell the number of subshells can be found by l = n -1. This means that for n = 1, the first shell, there is only l = 1-1 = 0 subshells.
