Y’all help me out i’m so dumb 3

Nitrogen monoxide and water react to form ammonia and oxygen, like this: 4no (g) +6h2o (g) →4nh3 (g) +5o2 (g) the reaction is en

Archy [21]

The concept used here is the Le Chatelier's principle. When a disturbance is introduced to the system, it favors the direction of reaction that minimizes the disturbance to regain equilibrium.

In endothermic reactions, the forward reaction is favored when the temperature is low. Otherwise, the reverse reaction is favored. When you add the amounts of substances on the reactant side, more products would formed favoring the forward reaction. If you increase concentration on the product side, you form more reactants so it would favor the reverse reaction. Lastly, since 10 moles of gases are needed in the reactant side, it would be favored during high pressure reaction.

3 0

3 years ago

9. A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent. A l

Arte-miy333 [17]

We have that the the liquid is

C_2H_5OH (ethanol
And at a condition of H_2SO4 as catalyst and temp 170

From the question we are told

A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent.
A low-boiling liquid was obtained instead of ethylene.
What was the liquid, and how might the reaction conditions be changed to give ethylene

<h3>Ethylene formation</h3>

Generally the equation is

2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20

Therefore

with ethanol at 140oC

The product is diethyl ethen

The reaction at 170 ethylene will give

C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)

Therefore

The the liquid is

C_2H_5OH (ethanol

For more information on Ethylene visit

brainly.com/question/20117360

8 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

How many neurons does atom in the model shown below have a.2 b.9 c.10 d.16

const2013 [10]

The atom in the model has 9 neurons.

7 0

3 years ago

Read 2 more answers

What is the percent yield of 20.0g P4 O10 are produced?

enyata [817]

Answer:

You must divide the grams of your actual yield by the grams of the theoretical yield and multiply by 100 in order to obtain percent yield

Explanation:

7 0

3 years ago