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nikklg [1K]
3 years ago
8

transition metal ions with partially filled D orbitals usually have color. based on your observations, which Solutions contain t

ransition-metal ions with partially filled d orbitals?
Chemistry
1 answer:
Yuliya22 [10]3 years ago
5 0
Since transition metals with partially filled d orbitals have color when in solution.  Therefore which ever solution has some color will likely contain a transition metal with a partially filled d orbital.
an example of this is solution with Cu²⁺ will have a blue tint to it.
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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

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3 years ago
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Which of the following is an inexhaustible energy resource?
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4 0
3 years ago
If 25.0 g NO are produced, how many grams of nitrogen gas are used?
mr Goodwill [35]

Using the stoichiometry of the reaction and the information provided in the question, the mass of N2 used is 11.62 g.

<h3>Chemical reaction</h3>

The term chemical reaction refers to the combiantion of two or substances to yiled one or more products. The reaction equation in this case is N2 + O2 --->2NO.

Now;

Number of moles of NO = 25g/30 g/mol = 0.83 moles

1 mole of N2 yields 2 moles of NO

x moles of N2 yileds 0.83 moles of NO

x = 0.415 moles

Mass of N2 = 0.415 moles * 28 g/mol = 11.62 g

Learn more about stoichiometry:

brainly.com/question/12166462

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Answer: B- a roll of photographic film

Explanation:

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