If the half-life of C-14 is 5730 years, how much C-14 would be left from a 5000 year-old sample that started as 150 grams?
1 answer:
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Answer 1) When a strong acid like 
reacts with 
usually the equilibrium shifts to the right because
As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when reacts with it generates carbon dioxide and water as a by product, if we are adding it will remove some of the 
molecule from the reaction mixture, which then tends to shift the equilibrium towards right.
Answer 2) The same would be observed in this case, if we replace with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.
As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing
and 
ions from the solution.
As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when
Answer 2) The same would be observed in this case, if we replace
As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing
Answer:
The change in the pH (ΔpH) is 2,17
Explanation:
The reaction:
CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻
<em>(1)</em>
In equilibrium, a solution of CH₃NH₂ 4,7M produces:
[CH₃NH₂] = 4,7 - x
[CH₃NH₃⁺] = x
[OH⁻] = x
Replacing in (1):
x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0
The solutions are:
x = -0,0456 No physical sense. There are not negative concentrations.
x = 0,04515 Real answer.
The concentration of [OH⁻] is 0,04515 M.
As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:
<em>pH = 12,65</em>
The addition of 6,7M produce this changes in concentrations:
[CH₃NH₂] = 4,656 + x
[CH₃NH₃⁺] = 6,74515 - x
[OH⁻] = 0,04515 - x
Replacing in (1) you will obtain:
x² - 6,7907x + 0,3025 = 0
Solving for x:
x = 6,74586 No physical sense
x = 0,04484 Real answer.
Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M
pOH = 3,51.
<em>pH = 10,49</em>
Thus ΔpH is 12,65 - 10,49 = <em>2,16 ≈ 2,17</em>
I hope it helps!