Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water
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Answer:
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Explanation:
Average rate of the reaction is defined as ratio of change in concentration of reactant with respect to given interval of time.
![R_{avg}=-\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
Where :
= initial concentration of reactant at 
.
= Final concentration of reactant at 
.
2A+3B → 3C+2D
![R_{avg}=-\frac{1}{2}\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
The concentration of A at (
) = 
The concentration of A at (
) = 
The average rate of reaction in terms of the disappearance of reactant A in an interval of 0 seconds to 20 seconds is :
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Answer:
An Element
Explanation:
An Element is made up of only one type of atom (Gold, Silver, etc.)
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Answer:
V₂ = 107.84 L
Explanation:
Given data:
Initial volume = 100 L
Initial pressure = 80 KPa (80/101 =0.79 atm)
Initial temperature = 200 K
Final temperature =273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm
V₂ =21567 atm.L.K /200 K.atm
V₂ = 107.84 L
Answer:
6.25%
Explanation:
Given data:
Half life of lutetium-117 = 6.75 days
Percentage remaining after 27 days = ?
Solution;
Number of half lives = Time elapsed / half life
Number of half lives = 27 days / 6.75 days
Number of half lives = 4
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At 3rd half life = 25%/2 = 12.5%
At 4th half life = 12.5%/2 = 6.25%
