The pressure of the gas = 40 atm
<h3>Further explanation</h3>
Given
200 ml container
P = 2 atm
final volume = 10 ml
Required
Final pressure
Solution
Boyle's Law
At a fixed temperature, the gas volume is inversely proportional to the pressure applied
Input the value :
P₂ = P₁V₁/V₂
P₂ = 2 x 200 / 10
P₂ = 40 atm
Answer:
55.85 grams of Fe is formed.
Explanation:
Identify the reaction:
2Fe₂O₃ + 3C → 4Fe + 3CO₂
Identify the limiting reactant, previously determine the mol of each reactant
(mass / molar mass)
10 g / 12 g/m = 0.83 moles C
80 g / 159.7 g /m = 0.500 moles Fe₂O₃
2 moles of oxide need 3 moles of C, to react
0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol
I have 0.83 moles of C, so C is the excess.
The limiting is the oxide.
3 mol of C need 2 mol of oxide to react
0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)
Ratio is 2:4 (double)
If I have 0.5 moles of oxide, I will produce the double, in the reaction.
1 mol of Fe, will be produce so its mass is 55.85 g
Answer:
Lowering the object near the ground decreases its <u>potential energy.</u>
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Explanation:
Potential Energy : Energy possessed by the object due to its shape ,Size and Position is called potential energy.
Example :
Change in shape and size : When you compress the spring , potential energy is introduced in it . So it expand quickly when you remove your hand.
Change in position : when you swing , after attaining maximum height (extreme ends) , the swing comes back on its on .This is because at maximum height ,the swing has<u> maximum Potential energy . </u>Hence it fall back on its on because it already has potential energy.
Potential energy(P) is given by the formula :
P = mgh
where ,
m= mass of the object
g = acceleration due to gravity
h = height of the object from the ground
If the height of the object increases from the ground , its potential energy also get increase.
<u><em>On lowering the object The height of the object from the ground reduces . So potential energy also reduces.</em></u>
