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 %
The only two that apply is Oxygen and Water. The last one would be Carbohydrates.
Answer:
Ksp = [ Cu+² ] [ OH-] ²
molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol
Ksp = [ Cu+² ] [ OH-] ²
Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰
|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|
|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|
<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|
|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>
Ksp = [ Cu+² ] [ OH-] ²
2.2 ×10-²⁰ = (S)(2S)²= 4S³
![s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}](https://tex.z-dn.net/?f=s%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B2.2%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2020%7D%20%7D%7B4%7D%20%7D%20%20%3D%201.8%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%207%7D%20)
S = 1.8 × 10-⁷ M
The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M
Solubility of Cu (OH)2 =

<h3>
Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>
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Answer:
the number of carbon and hydrogen atoms present in the molecule of Ethane that is it contains two carbon atoms and six hydrogen atoms
Explanation: