Molar mass of h2co3
Total molar mass for hydrogen atoms
Total mass for carbon atoms
Total molar mass for oxygen atoms
Hydrogen percent composition
Oxygen percent composition

1. molar mass of h2co3: 62.03 g

2. total molar mass for hydrogen atoms: 2.02 g

3. total molar mass for carbon atoms: 12.01 g

4. total molar mass for oxygen atoms: 48.00 g

5. hydrogen percent composition: 3.3%

6. oxygen percent composition: 77.4%

Explanation:

1. 1.01+1.01+12.01+16.00+16.00+16.00= 62.03

2. 1.01×2= 2.02

3. 12.01×1= 12.01

4. 16.00×3= 48.00

5. (2.02/62.03)×100= 3.3%

6. (48.00/62.03)×100= 77.4%

(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

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o