Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
<span>Begin by classifying which energy level, and indirectly principal quantum number, n, resembles to the N shell.
no. of orbitals =n2
In your case, the fourth energy level will contain
n=4⇒no. of orbitals= 4^2=16
The number of subshells is given by the principal quantum number.
no. of subshells=n
In your case, the fourth energy level will have
no. of subshells = 4 this is the answer
to check:
the fourth energy shell will can hold a thoroughgoing of no. of electrons=2⋅42=32 e−</span>
Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
Answer:
there are 2.2 moles of Li in 15 grams of Li
