Chemistry, 10.01.2020 06:31 welathyasf

A34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of water. what is a possible molecular formula of the hydro-carbon?

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Answer from: riveranikki2

The answer to your question is CH₄

Explanation:

Process

1.- Write the possible chemical reaction

CxHy + O₂ ⇒ CO₂ + H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

44 g of CO₂ 12 g of carbon

93.5 g of CO₂ x

x = (93.5 x 12) / 44

x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

18 g of water 2 g of hydrogen

76.5 g of water x g of hydrogen

x = (76.5 x 2) / 18

x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

12g of carbon 1 mol

25.5g of carbon x

x = (25.5 x 1) / 12

x = 2.12 moles of carbon

1 g of hydrogen 1 mol

8.5 g of hydrogen x

x = (8.5 x 1) / 1

x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

CH₄

Answer from: Quest

Noble gases is far right

Answer from: Quest

(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

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A34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon...

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