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
Answer:
H2O(s)
Explanation:
because it is a solid and an aqueous can not conduct electricity
You need to find the mole of glucose by using the formula n = m/Mr. Once you have found the mole you need to follow the stoichiometric process by unknown/known using the co-efficients. Then you need to multiple your answer by the known mole of glucose whoch you have previously calculated. After that you should get the mole of oxygen then you just need to transpose the equation to get m = n×Mr, substitute ypur answers in, remebering you can find your Mr via the period table, you have your answer
The answer should be 261.97 g/mol
It's 3. There are two Cl in the reactants and 3 in the products, and they need to be balanced/equal. Only way to do that is add three on the left side, which will give you 6 Cl (3×2Cl) and then 2 on the other side, which will also give you 6 Cl (2×3Cl)
