How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
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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!
Answer: The percent yield is, 93.4%
Explanation:
First we have to calculate the moles of Na.
Now we have to calculate the moles of
The balanced chemical reaction is,
As, 1 mole of bromine react with = 2 moles of Sodium
So, 0.189 moles of bromine react with = moles of Sodium
Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.
As, 1 mole of bromine give = 2 moles of Sodium bromide
So, 0.189 moles of bromine give = moles of Sodium bromide
Now we have to calculate the percent yield of reaction
Therefore, the percent yield is, 93.4%
Answer:
<h2>3.92 g/cm³</h2>Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>3.92 g/cm³</h3>Hope this helps you