Answer:
The activation energy
Explanation:
The activation energy is the energy hump that lies between reactants and products. It is the energy barrier that reactants must cross before they are converted into products.
Based on the collision theory, only particles that possess the activation energy are able to collide in such a way that leads to reaction.
Collision of particles having an energy content less than the activation energy of the reaction merely leads to elastic collision between such particles.
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>
I hope I helped you^_^
Answer:
Percent by mass of water is 56%
Explanation:
First of all calculate the mass of hydrated compound as,
Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g
Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g
Mass of Oxygen = O × 14 = 16 × 14 = 224 g
Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g
Mass of Na₂S0₄.10H₂O = 322.24 g
Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...
Mass of water = 10 × 18.02
Mass of water = 180.2 g
Now, we will apply following formula to find percent of water in hydrated compound,
%H₂O = Mass of H₂O / Mass of Hydrated Compound × 100
Putting values,
%H₂O = 180.2 g / 322.24 g × 100
%H₂O = 55.92 % ≈ 56%
Combustion reaction
Key: O2
O2 is normally in a chemical formula when you are used to burn anything, so basically, anything with O2 involves burning.
Answer:
<h3>I don't know what is the answer of your question sorry never mind..</h3>
Explanation:
<h3>And please marks me as brainliest... </h3>