Answer:
73.4% is the percent yield
Explanation:
2KClO₃ → 2KCl + 3O₂
This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.
We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃
In the theoretical yield of the reaction we say:
2 moles of potassium chlorate can produce 3 moles of oxygen
Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂
The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g
But, we have produced 115 g. Let's determine the percent yield of reaction
Percent yield = (Produced yield/Theoretical yield) . 100
(115g / 156.6g) . 100 = 73.4 %
Conductivity is a physical property
Odor is also a physical property
Answer: Decreases across the period and increases down the group
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
.</span>
Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction
Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.
