<span> the line-emission spectrum of an atom is caused by the energies released when electrons. releases energy of only certain values. </span>
The reaction is:
2 KClO3(s) → 3 O2(g) + 2 KCl(s) <span>
<span>A catalyst simply lowers the activation energy so MnO2 is not
part of the overall reaction.
By stoichiometry:
<span>3.45 g KClO3 x 1 mol/ 122.55g KClO3 x 3 mol O2/ 2 mol KClO3 x
31.99g/ 1 mol O2 = 331.096/ 245.1 = 1.35 grams O2 produced
Answer:1.35 grams O2</span></span></span>
Answer: pH = 4.996
Explanation:
No of moles = molarity x volume
:• no of moles of CH3COOH = 0.1M x 0.1L
n(CH3COOH) = 0.1mol
Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.
Therefore, Moles of CH3COOH becomes,
0.1 - 0.03 = 0.07 mol
Subsequently, the moles of CH3COONa increases and becomes,
0.08 + 0.03 = 0.11 mol
Using the Hendersom-Hasselbach equation,
pH = pKa + log [Moles of conjugate÷ moles of Ch3COOH]
From literature, pKa of Ch3COOH is 4.8
Thus,
pH = 4.8 + log [0.11/0.07]
pH = 4.8 + 0.1963
pH = 4.996
Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Explanation:
Given that;
equilibrium constant Kc = 1.29 × 10⁻² at 600k
the equilibrium concentrations of reactant and products = ?
when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]
Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M
COCl₂(g) ----------> CO(g) + Cl₂(g)
0.280 0 0 ------------ Initial
-x x x
(0.280 - x) x x ----------- equilibrium
we know that; solid does not take part in equilibrium constant expression
so
KC = [CO][Cl₂] / COCl₂
we substitute
1.29 × 10⁻² = x² / (0.280 - x)
0.0129 (0.280 - x) = x²
x² = 0.003612 - 0.0129x
x² + 0.0129x - 0.003612 = 0
x = -b±√(b² - 4ac) / 2a
we substitute
x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]
x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2
x = [-0.0129 ± 0.1209] / 2
Acceptable value of x =[ -0.0129 + 0.1209] / 2
x = 0.108 / 2
x = 0.054
At equilibrium
[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
9moles of O₂
Explanation:
Given parameters:
Number of moles of KClO₃ = 6moles
Unknown:
Number of moles of O₂ = ?
Solution:
KClO₃ is often known to undergo a decomposition reaction to produce KCl and O₂.
2KClO₃ → 2KCl + 3O₂
From the given equation, we know;
2 moles of KClO₃ will produce 3 mole of O₂
6 mole of KClO₃ will produce 
= 9moles of O₂
learn more:
Number of moles brainly.com/question/1841136
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