Answer:

Explanation:
Hello,
In this case, as the atomic mass of coppe is 63.546 g/mol, with the given mass with can compute the moles as shown below:

Best regards.
Answer:
Palms and other small trees
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
Explanation:
substance Q could be <em><u>oxygen (O2)</u></em>
substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>
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.