Vanadium(V) oxide is the inorganic compound with the formula V₂O₅. Commonly known as vanadium pentoxide, it is a brown/yellow solid, although when freshly precipitated from aqueous solution, its colour is deep orange. Because of its high oxidation state, it is both an amphoteric oxide and an oxidizing agent.
formula:V2O5
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Answer:
A. It would float with about 80% of the cube below the surface of the water and 20% above the surface.
Explanation:
The choice that best describes what happens to cube of the given density value is that it would float with about 80% of the cube would be below the surface of the water and 20% above the surface.
Density is the mass per unit volume of a substance. The more mass a body has relative to volume, the great it's density. In short, density is directly proportional to mass and inversely related to volume.
The density of water is 1g/mL
If the density of the cube were to be the same with that of water, the substance will just mix up with water .
Here the density is less than that of water.
The density is 0.2g/mL
Therefore, 20% will stay afloat and 80% will be below the surface of the water.
Using the Rydberg formula, the spectral line of H - atom is suitable for this purpose is Paschen, ∞ → 3.
- Using the Rydberg formula;
1/λ = RH(1/nf^2 - 1/ni^2)
Given that;
λ = wavelength
RH = Rydberg constant
nf = final state
ni = initial state
- When final state = 3 and initial state = ∞
Then;
1/λ = 1 × 10^7 m-1 (1/3^2 - 1/ ∞^2)
1/λ = 1 × 10^7 m-1 (1/3^2 )
λ = 900 nm
Hence, the correct answer is Paschen, ∞ → 3
Learn more about the Rydberg formula; brainly.com/question/17753747
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
