Ammonia react with phosphorus acid to form a compound that contains 28.2% of nitrogen 20.8%, phosphorus 8.1% of hydrogen 42.9%oxygen. calculate the empirical formula of this compound

Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

Explanation:

This is the reaction:

NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O

In 100 g of compound we have:

28.2 g N

20.8 g of P

8.1 g of H

42.9 g of O

Now we divide each  between the molar mass:

28.2 g / 14 g/mol = 2.01 mol

20.8 g / 30.97 g/mol = 0.671 mol

8.1 g / 1 g/mol = 8.1 mol

42.9 g / 16 g/mol =  2.68 mol

And we divide again between the lowest value of moles

2.01 mol / 0.671 mol → 3

0.671 mol / 0.671 mol → 1

8.1 mol / 0.671 mol → 12

2.68 mol / 0.671 mol → 4

Molecular formula will be: N₃PH₁₂O₄ → (NH₄)₃PO₄

Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

with this one i know the answer in craters one reason i looked it up and i learned about it there is my evidence

(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