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prohojiy [21]
3 years ago
10

What is the standard unit that astronomers use to measure the size of a star

Chemistry
1 answer:
g100num [7]3 years ago
7 0
Parsecs or abbreviated pc
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Which states of matter only appear in the hydrosphere
Oliga [24]
I would say the answer is liquids
7 0
2 years ago
Would you expect the wavelength of maximum absorbance for [Cu(NH3)4]2+ to be greater than or less than the wavelength of maximum
DENIUS [597]

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.

7 0
3 years ago
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
What is 125000 in scientific notation?
Usimov [2.4K]

Answer:

1.25 x 10^5

Explanation:

7 0
3 years ago
How do you determine the volume of a regular object if you don't know the density?
Ann [662]

Answer:

mol times or devided by molar volume

4 0
3 years ago
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