How many particles are in 35 g of Fe2(SO4)3? Round your answer to 2 decimal places. Help! 80 points

First, you need to know the molar mass of the compound. The compound Aluminum Sulfate has a molar mass of 342.15 g/mol. From the subscript of the compound, there are 4*3 = 12 moles of oxygen in 1 mole of the compound. The solution is as follows:

6.90 g Al₂(SO₄)₃*(1 mol/342.15 g)*(12 mol O/1 mol Al₂(SO₄)₃)*(6.022×10²³ atoms/mole) = 1.457×10²³ atoms of oxgyen

Explanation:

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(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