The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
given that :
molarity = 2.16 M
volume = 250 mL = 0.25 L
the molarity is given as :
molarity = number of moles / volumes in L
from this we can calculate the number of moles, we get :
number of moles of K₂Cr₂O₇ = molarity × volume
number of moles of K₂Cr₂O₇ = 2.16 × 0.25
number of moles of K₂Cr₂O₇ = 0.54 mol
Thus, The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
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Answer:
D
Explanation:
When lead ions and sulfate ions bond, they form sediment so neither a nor b can be the answer.
The important thing is that two nitrate ions were originally bonded with one lead ion, while two potassium ions bonded with a sulfate ion.
Finally, since potassium and nitrate ions don't form sediment these two ions must remain. Therefore the answer is D
Answer:
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Explanation:
v8y to c g. fu uf gu t8 g8. g8 u g 8 g gi gi gi g gi if fu%:/"
Answer:
1.64x10⁻¹⁸ J
Explanation:
By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.
When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:
E = hc/λ
Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.
The wavelength can be calculated by:
1/λ = R*(1/nf² - 1/ni²)
Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:
1/λ = 1.097x10⁷ *(1/1² - 1/2²)
1/λ = 8.227x10⁶
λ = 1.215x10⁻⁷ m
So, the energy is:
E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)
E = 1.64x10⁻¹⁸ J
Answer:
this is what i got
Explanation:
α-decay: When a radioactive nucleus disintegrates by emitting an αα-particle, the atomic number decreases by two and mass number decreases by four. Example: 88Ra226→86Rn222+2He4.
