Answer: 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.
Explanation:
The balanced chemical equation is:
To calculate the moles, we use the equation:
moles of manganese = 
According to stoichiometry :
3 moles of
is produced by = 3 moles of
Thus 20.4 moles of
is produced by =
of
Mass of
(1kg=1000g)
Thus 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.
Answer:

Explanation:
Percent yield is a ratio of the actual yield to the theoretical yield. It is found using this formula:

The actual yield is 12 liters, because that was actually produced in the lab.
The theoretical yield is 20 liters, because that was the expected yield.



For this reaction, the percent yield is 60%.
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Answer:

Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure

(c) Convert the pressure to atmospheres

Answers: -
For high kinetic energy, the object must have high speed of movement.
1) An airplane has a lot of kinetic energy. Airplanes move at high speed and thus posses a lot of kinetic energy.
2) A bullet from a gun has a lot of kinetic energy due to the high speed of bullet.
3) A formula one car moving at high speeds have a lot of kinetic energy.
4) A train moving at high speed has lots of kinetic energy.
5) An asteroid has a lot of kinetic energy due to it's high speed.
6) A roller coaster moving at high speeds have a lot of kinetic energy.
7) A missile fired from a fighter plane has lots of kinetic energy.