Ammonia and oxygen react to form nitrogen monoxide and water, like this:

1 answer:

6 0

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.

You might be interested in

Magma can partially crystallize at depth and then rise to shallow depths where the remaining magma solidifies. The early-formed

Doss [256]

<h2>Phenocrysts and Porphyritic texture </h2>

Explanation:

The early formed crystals are of phenocrysts and the texture of these crystals is porphyritic texture.
This crystallization occurs when early-forming plagioclase crystals which are rich in calcium start coating with plagioclase crystals which are rich in sodium.
On cooling, the magma is then processed in a volcanic eruption, after the eruption the liquid which is left behind will start cooling and forms a porphyritic texture.

5 0

2 years ago

Use the formula: PV=nRT<br><br> If P=12, V=22.4, R=.0821 and T=100, then what would n=?

Lina20 [59]

$PV=nRT \\ (12)(22.4)=n(.0821)(100) \\ 268.8=8.21n \\ 268.8/8.21=8.21n/8.21 \\ n=32.74$
n ≈ 32.74
Hope this helps!

7 0

2 years ago

What equation represents the pressure of a gas under ideal conditions?

Leno4ka [110]

P=nRTV
hope this help<span />

8 0

3 years ago

Read 2 more answers

1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced

koban [17]

Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$

3 0

2 years ago

How does increasing the temperature affect collisions?

ddd [48]

Combustion Have a great day!

8 0

2 years ago

Read 2 more answers