Mass of ammonia produced : 121.38 g
<h3>Further explanation</h3>
Given
Reaction
3H₂(g) + N₂(g) ⇒ 2NH₃(g)
100g of N₂
Required
Ammonia produced
Solution
mol of N₂ :
From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :
mass of NH₃(MW=17 g/mol) :
<span>Hydrogen carbon and oxygen commonly form covalent bonds.
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HOPE THIS HELPS!
Answer:
The answer is "Greater than zero, and greater than the rate of the reverse reaction".
Explanation:
It applies a rate of reaction to the balance, a forward response dominates until it reaches a constant. This process is balanced before 52 mmol of the reactant
, to which 3 is added. In balance, that rate of the forward reaction was its rate with forwarding reaction, both of which are higher than 0 as the response has achieved balance so that both species get a level greater than 0.
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
Answer:
<em>Hello, Your answer will be </em><em>B) Jacob's backyard is on the north side of his house.</em>
<em>Hope That Helps!</em>
