Mass = moles x molar mass
so mass of 6 moles of h2 is: 6×1×2 = 12g
KauCl4 :
K = + 1
au = + 7
Cl = - 2
hope this helps!
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)
Answer to this question is C. Regarding the volume.
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
