QUESTION 21 The combustion of ammonia in the presence of excess oxygen yields NO 2 and H 2O: 4 NH 3 (g) 7 O 2 (g) 4 NO 2 (g) 6 H 2O (g) The combustion of 14.4 g of ammonia consumes g of oxygen.

The answer to your question is 47.44 g of Oxygen

Explanation:

Data

mass of Ammonia = 14.4 g

mass of Oxygen = ?

Balanced chemical reaction

                 4NH₃  +  7O₂  ⇒  4NO₂  +  6H₂O

Process

1.- Calculate the molar mass of Ammonia

NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g

2.- Calculate the molar mass of Oxygen

O₂ = 7[16 x 2] = 7[32] = 224 g

3.- Use proportions to calculate the mass of Oxygen

                     68g of NH₃ 224 g of O₂

                      14.4 g of NH₃  x

                       x = (14.4 x 224) / 68

                       x = 3225.6/ 68

                       x = 47.44 g

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