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Archy [21]
3 years ago
6

Complexes containing metals with d10 electron configurations are typically colorless because ________. Complexes containing meta

ls with d10 electron configurations are typically colorless because ________. d electrons must be emitted by the complex in order for it to appear colored there are no d electrons to form bonds to ligands a complex must be charged to be colored there is no d electron that can be promoted via the absorption of visible light the empty d orbitals absorb all of the visible wavelengths
Chemistry
1 answer:
algol [13]3 years ago
8 0

Answer:

there is no d electron that can be promoted via the absorption of visible light

Explanation:

One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.

The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.

For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.

Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.

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Answer:

100 HZ 1,000 HZ 10,000 HZ there you go :)

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Why noble gas neon is an unreactive element?
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This is because it has a full outer valence shell! so there are 8 electrons and that means it doesn't have the urge the gain anymore 
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If 1.10 g of steam at 100.0 °C condenses into 38.5 g of water, initially at 27.0 °C, in an insulated container, what is the fina
Gwar [14]

Answer:

Temperature = 44.02°C

Explanation:

Insulated container indicates no heat loss to the surroundings.

The specific heat capacity of a substance is a physical property of matter. It is defined as the amount of heat that is to be supplied to a unit mass of the material to produce a unit change in its temperature.

The SI unit of specific heat is joule per kelvin and kilogram, J/(K kg).

Now,

Specific heat for water is 4.1813 Jg⁻¹K⁻¹.

Latent heat of vaporization of water is 2257 Jg⁻¹.

Energy lost by steam in it's process of conversion to water, is the energy acquired by water resulting in an increase in it's temperature.

Q = mS \Delta T

Q= Heat transferred

m= mass of the substance

T= temperature

Also,

Q = mL

L= Latent heat of fusion/ vaporization ( during phase change)

Now applying the above equations to the problem:

m_{w} S_{w} (T-27) = (m_{s} L) + m_{s}S_{w} (100 -T)

38.5 \times 4.1813 \times (T-27) = (1.10 \times 2257) + 1.10 \times 4.1813 \times (100 -T)

Temperature = 44.02°C

8 0
4 years ago
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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Explanation:

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