The solution would be like this for this specific problem:
First, we need to write
out the balanced reaction equation for the problem:<span>
"gaseous nitrogen with gaseous hydrogen to produce
gaseous ammonia"
gaseous nitrogen + gaseous hydrogen = gaseous ammonia
gaseous nitrogen = N2(g)
gaseous hydrogen = H2(g)
gaseous ammonia = NH3(g)
In here, I’m going to show you how to balance this from start
to finish:
N2(g) + H2(g) ↔ NH3(g) (basic, unbalanced equation that shows
reactants and product)
N2(g) + H2(g) ↔ 2NH3(g) (balances the nitrogens)
N2(g) + 3H2(g) ↔ 2NH3(g) (balances the hydrogens) =>
Balanced Equation
<span>Next, we write the Keq expression of the balanced reaction:
Keq = (∏[products]^n) /(∏[reactants]^n)
Keq = [NH3]^2 / [[N2][H2]]^3
Therefore, the correct
equilibrium constant expression is Keq = (NH3)2 / (N2)(H2)3.</span></span>
Its D for plato.
by other people who asked
Answer:
A symbiotic relationship.
I hope this helped! Brainliest would be greatly appreciated.
<h2>
Answer: 1.22 × 10⁻¹ m</h2>
<h3>Explanation:</h3>
Wavelength = Speed of Light ÷ Frequency
= (2.99 × 10⁸ m/s) ÷ ( 2.45 × 10⁹ /s)
= 1.22 × 10⁻¹ m
The wavelength of a light of frequency 2.45 × 10⁹ /s is 1.22 × 10⁻¹ m.
<u>Notes:</u>
Hz ≡ /s
<span>Answer:
To determine hybridization you count the # of un-bonded PAIR of electrons. Theny you count bonded domains (a double bond still counts as one bonded domain, so does a triple bond).
For the first Carbon on top, you see it bonded to 2 oxygens and 1 carbon.However, when you count up the valence electrons that would be present, there is supposed to be 2 more electrons (ONE electron pair) on carbon. To do hybridization you must also be familiar with drawing lewis structures. So when you draw the pair of un-bonded electrons on carbon, you see that there is ONE un-bonded electron pair, and THREE bonded domains.
ONE plus THREE = FOUR.
so that would be sp3.</span>