Answer:
Can you send the picture of this question
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
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To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
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Answer:co2
Explanation:because of the oxygen levels
Answer:
22.73s
Explanation:
The reaction is a second order reaction, we know this by observing the unit of the slope.
rate constant = k = 0.056 M-1s-1
the initial concentration of BrO- [A]o = 0.80 M
time = ?
Final concentration [A]t= one-half of 0.80 M = 0.40M
1 / [A]t = kt + 1 / [A]o
1 / 0.40 = 0.056 * t + 1 / 0.80
t = (2.5 - 1.25) / 0.056
t = 22.73s