Answer:
After the transfer the pressure inside the 20 L vessel is 0.6 atm.
Explanation:
Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:
Therefore, for this problem the step by step explanation is:
Clearing P2 and replacing
Different isotopes of the same element must have a different mass number
Explanation:
Moles of NaOH = 10g / (40g/mol) = 0.25mol.
0.25mol / 500g = 0.50mol / 1000g = 0.50mol/dm³.
The molarity is 0.50mol/dm³.
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
