X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160
so, moles = 85.2 / (4y+160)
Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0
so, 48 = 16 x 10 x [85.2 / (4y+160) ]
Solve the equation to get y.
y = 31
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
Answer:
1.3 meters
Explanation: use newton third law equation.
Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
