Answer:
Explanation:
Hello,
In this case, by knowing the given reference reactions, one could rearrange them as follows:
Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):
Consequently, the equilibrium constant is computed as:
![Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%20%2A%20%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%5E2%7D%20%3DKp_2%2AKp_3%3D4.35x10%5E%7B18%7D%2A7.056x10%5E%7B-13%7D%3D3.07x10%5E6)
Best regards.
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
The half life of uranium- 238 is
4.46 x10^9 years
Explanation
Half life is the time taken for the radioactivity of a isotope to fall to half its original value.
The original mass of uranium-238 is 4.0 mg
Half of original mass of uranium = 4.0 mg /2 = 2.0 mg
since it take 4.46 x 10^ 9 years for the sample to half the half life of uranium -238 = 4.46 x10^9 years
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