Answer:
The correct answer is: 1.035 x 10⁻³ M
Explanation:
The dissociation equilibrium for acetic acid (CH₃COOH) is the following:
CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵
The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):
![Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7D%20COO%5E%7B-%7D%20%5D%5BH%5E%7B%2B%7D%20%5D%7D%7B%5BCH_%7B3%7D%20COOH%5D%7D%3D%201.8%20x%2010%5E%7B-5%7D)
Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:
![\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.016%20M%29%5BH%5E%7B%2B%7D%20%5D%7D%7B%280.92M%29%7D%3D%201.8%20x%2010%5E%7B-5%7D)
⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M
By hot springs, fumaroles and geysers.
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Answer:N
2
+ 3
H
2
-----> 2N
H
3
Explanation:
N
2
+
H
2
-----> N
H
3
Let us balance this equation by counting the number of atoms on both sides of the arrow.
N
2
+
H
2
-----> N
H
3
N=2 , H=2 N=1, H=3
To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N
H
3
on RHS
N
2
+
H
2
-----> 2N
H
3
N=2 , H=2 N=2 , H= 6
To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of
H
2
on LHS
N
2
+ 3
H
2
-----> 2N
H
3
N=2 , H=6 N=2 , H= 6
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