The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>
Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :
Fe O
2 3 is what i would put
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
Answer:
a) 1.866 × 10 ⁻¹⁹ J b) 3.685 × 10⁻¹⁹ J
Explanation:
the constants involved are
h ( Planck constant) = 6.626 × 10⁻³⁴ m² kg/s
Me of electron = 9.109 × 10 ⁻³¹ kg
speed of light = 3.0 × 10 ⁸ m/s
a) the Ek ( kinetic energy of the dislodged electron) = 0.5 mu²
Ek = 0.5 × 9.109 × 10⁻³¹ × ( 6.40 × 10⁵ )² = 1.866 × 10 ⁻¹⁹ J
b) Φ ( minimum energy needed to dislodge the electron ) can be calculated by this formula
hv = Φ + Ek
where Ek = 1.866 × 10 ⁻¹⁹ J
v ( threshold frequency ) = c / λ where c is the speed of light and λ is the wavelength of light = 358.1 nm = 3.581 × 10⁻⁷ m
v = ( 3.0 × 10 ⁸ m/s ) / (3.581 × 10⁻⁷ m ) = 8.378 × 10¹⁴ s⁻¹
hv = 6.626 × 10⁻³⁴ m² kg/s × 8.378 × 10¹⁴ s⁻¹ = 5.551 × 10⁻¹⁹ J
5.551 × 10⁻¹⁹ J = 1.866 × 10 ⁻¹⁹ J + Φ
Φ = 5.551 × 10⁻¹⁹ J - 1.866 × 10 ⁻¹⁹ J = 3.685 × 10⁻¹⁹ J
Answer:
Explanation:
Let us first take a look at the image below;
In the acid - base reaction; we can see the transfer of electrons that takes place;
We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.
The stronger acid is shown with the one with more negative 
Value.
∴ The equilibrium constant for the acid-base reaction is expressed as:
From Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.
From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:
