Dinitrogen monoxide gas decomposes to form nitrogen gas and oxygen gas. how many grams of oxygen are formed when 10.0 g of dinitrogen monoxide decomposes?

There is formed 3.6352 grams of oxygen

Explanation:

Step 1: The balanced equation

2N2O(g)⇔2N2(g)+O2(g)

Step 2: Calculate moles of dinitrogen monoxide

Moles = mass/ Molar mass

Moles of N2O = 10 grams of N2O / 44.01 g/mole

Moles of N2O = 0.227 moles

Step 3: Calculate moles of oxygen

When 2 moles of N2O is consumed, there is produced 1 mole of O2 ( and 2 moles of N2)

That means for 0.2272 moles of N2O consumed, there is produced 0.1136 moles of O2

Step 4: Calculate mass of O2

mass of O2 = moles of O2 * Molar mass of O2

mass of O2 = 0.1136 moles * 32 g/moles

mass of O2 = 3.6352 grams

There is formed 3.6352 grams of oxygen

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

let that person have a turn to speak : )