An unknown oxide of mercury decomposes when heated to form mercury metal and oxygen gas. when a 1.048 g sample of this unknown is heated, 0.971 g of mercury remains.

a. calculate the moles of mercury and moles of oxygen in the compound.

b. what is the empirical formula of this oxide?

2. a side reaction in today

Moles of mercury = 0.00484 mol.

Moles of oxygen = 0.00481 mol.

The empirical formula is =

Explanation:

Mass of mercury = 0.971 g

Molar mass of mercury = 200.59 g/mol

Moles of mercury = = 0.00484 mol.

Since, the oxide contains only mercury and oxygen. SO,

Mass of O in the sample = Total mass - Mass of Hg

Mass of the sample = 1.048 g

Mass of O in sample = 1.048 g - 0.971 g = 0.077 g

Molar mass of oxygen = 15.999 g/mol

Moles of oxygen = = 0.00481 mol.

Taking the simplest ratio for Hg and O as:

0.00484 : 0.00481

= 1 : 1

The empirical formula is =

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"you didn't chose the stud life. the stud life chose you."

(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