When 1.375 g of copper (ii) oxide is reduced on heating in a current of hydrogen, the weight of copper remaining after the reaction is complete is 1.098 g. what is the percent composition of oxygen?

20.11% of oxygen

Explanation:

The reaction is:

2CuO ⇒ 2Cu + O2

If you have an initial amoung of a sample, thi may be have impurities, so you have to start from products to know what you had at the begining.

to make a relation between reagents and products you need to change the mass to mol, then you can relate them with the molar ratio.

Now you have the initial amount of oxygen, to convert to percent you must divide the mass of oxygen between the mass of the sample and multiply by 100

(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