Moles = (6.74*10^23)/(6.02*10^23) =1.119 moles
1.119*44.09=49.36g
(80+125+45) / 10 = 250/10 =25
25 meters per minute= 0.41 meters/second
the direction and stopping time is irrelevant to the problem.
1 mole C3H8 produces 4 moles H2O. So, first we convert 32 grams of propane to moles and then find moles of H2O. Then convert moles of H2O to grams of H2O
Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O
Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O
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
One side will burn to a crisp and the other side will freeze... and after some time the earth would be pulled into the sun and cooked
