How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and 0'C?

26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and

Explanation:

According to ideal gas equation:

P = pressure of gas = 901 mm Hg = 1.18 atm   (760 mm Hg = 1 atm)

V = Volume of gas = 5.00 L

n = number of moles = ?

R = gas constant =

T =temperature =

The chemical reaction for the decomposition of calcium carbonate follows the equation:

According to stoichiometry:

1 mole of carbon dioxide is produced by = 1 mole of calcium carbonate

Thus 0.263 moles of carbon dioxide is produced by = moles of calcium carbonate

Mass of calcium carbonate=

26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied