Chemistry, 19.03.2020 08:01 leomessifanboy678

Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reaction finds that liters per second of dioxygen are consumed when the reaction is run at and . Calculate the rate at which nitrogen monoxide is being produced. Give your answer in kilograms per second. Round your answer to significant digits.

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Answer from: camiee13dvivvj

The given question is incomplete. The complete question is as follows.

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is the oxidation of ammonia. In this reaction, gaseous ammonia reacts with dioxygen gas to produce nitrogen monoxide gas and water.

Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reaction finds that 645. liters per second of dioxygen are consumed when the reaction is run at 195.oC and 0.88 atm. Calculate the rate at which nitrogen monoxide is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.

Explanation:

Chemical equation for the oxidation of ammonia is as follows.

$4NH_{3}(g) + 5O_{2}(g) \rightleftharpoons 4NO(g) + 6H_{2}O(l)$

Then volume of $O_{2}$ per second consumed is as follows.

V = $\frac{645 L}{1 sec}$

As this reaction is taking place at a temperature of $195^{o}C$ (468.15 K) and pressure 0.88 atm. Hence, moles of consumption of $O_{2}$ are calculated as follows.

n = $\frac{PV}{RT}$

= $\frac{0.88 atm \times 645 L/sec}{0.08206 L atm/mol K \times 468.15 K}$

= 14.77 mol $O_{2}$ /sec

When 5 moles of $O_{2}$ produces 4 moles of NO then the amount of NO produced from 14.77 mol

$14.77 mol O_{2} /sec \times \frac{4 mol NO}{5 mol O_{2}} \times \frac{30.01 g NO}{mol NO}$

= 354.60 g/s

Therefore, NO formed per second is as follows.

$354.60 g/s \times \frac{kg}{10^{3} g}$

= 0.35 kg/s

Thus, we can conclude that the rate at which nitrogen monoxide is being produced is 0.35 kg/s.

Answer from: Quest

10 years is the answer

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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