2.23 moles of propane react when 294 g of CO₂ is formed .
<h3>What is moles ?</h3>
Moles is a unit which is equal to the molar mass of an element.
A reaction is given
C₃H₈ +50₂ → 3CO₂ + 4H₂O
Grams of CO₂ formed = 294 gm 
In moles = 294 /44 = 6.68 moles.
Let x be the moles of C₃H₈ is x 
Mole ratio of CO₂ to C₃H₈ = 3 : 1
so
6.68 /x = 3/1
x = 6.68 /3 = 2.23 moles
Therefore 2.23 moles of propane react when 294 g of CO₂ is formed .
To know more about Moles
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Answer:
3.6mol Li
Explanation:
To get the amount of moles from the mass we divide the mass by the molar mass. 
25g ÷ 6.941g/mol = 3.6mol Li
 
        
             
        
        
        
2.<span>Two or more atoms must collide, with proper orientation, with energy greater than or equal to the activation energy for a reaction to occur</span>
        
                    
             
        
        
        
Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1,  [conjugate base] = [acid] in solution. 
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!
 
        
             
        
        
        
<span>The answer to this would be radioactive decay. This is also known as "nuclear decay" or "radioactivity".</span>