Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml
Answer:
2.1 kg of water
Explanation:
Step 1: Given data
- Moles of lithium bromide (solute): 4.3 moles
- Molality of the solution (m): 2.05 m (2.05 mol/kg)
- Mass of water (solvent): ?
Step 2: Calculate the mass of water required
Molality is equal to the moles of solute divided by the kilograms of solvent.
m = moles of solute/kilograms of solvent
kilograms of solvent = moles of solute/m
kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg
A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
Oceans are the most common of course
