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Verdich [7]
3 years ago
5

Two elements in the second transition series (y through cd ) have three unpaired electrons in their 3+ ions. what elements fit t

his description?

Chemistry
2 answers:
Tanzania [10]3 years ago
7 0
Molybdenum and Rhodium, I believe. All of the elements in these 2 groups follow unusual orbital filling rules.

Hope this helped!
kotykmax [81]3 years ago
4 0

Answer: The elements that have 3 unpaired electrons in their +3 oxidation states are Molybdenum and Palladium.

Explanation: The elements in the second transition series are:

Yttrium (Y), Zirconium (Zr), Niobium (Nb), Molybdenum (Mo), Technetium (Tc), Ruthenium (Ru), Rhodium (Rh), Palladium (Pd), Silver (Ag) and Cadmium (Cd).

The electronic configuration of these elements are:

Y=[Kr]5s^24d^1             ;      Y^{3+}=[Kr]

Zr=[Kr]5s^24d^2           ;      Zr^{3+}=[Kr]4d^1

Nb=[Kr]5s^14d^4           ;      Nb^{3+}=[Kr]4d^2

Mo=[Kr]5s^14d^5           ;     Mo^{3+}=[Kr]4d^3

Tc=[Kr]5s^14d^6            ;     Tc^{3+}=[Kr]4ds^4

Ru=[Kr]5s^14d^7            ;     Ru^{3+}=[Kr]4d^5

Rh=[Kr]5s^14d^8            ;     Rh^{3+}=[Kr]4d^6

Pd=[Kr]5s^04d^{10}          ;     Pd^{3+}=[Kr]4d^7

Ag=[Kr]5s^14d^{10}           ;     Ag^{3+}=[Kr]4d^8

Cd=[Kr]5s^24d^{10}          ;     Cd^{3+}=[Kr]4d^9

From the above configurations of the elements in their +3 ionic state, the elements having 3 unpaired electrons are Molybdenum and Palladium.

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muminat

Answer:

the three products of destructive distillation of coal are:-

1. coal gas

2. coke

3. ammonia liquor

7 0
3 years ago
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Write the formulas for the ionic compounds formed by the following:
PSYCHO15rus [73]

1. KI

2. AlBr₃

3. CsNO₃

4. Al₂(CO₃)₃

Explanation:

1. potassium (K⁺) iodine (I⁻) - KI

2. aluminium (Al³⁺) bromine (Br⁻) - AlBr₃

3. caesium (Cs⁺) nitrate (NO₃⁻) - CsNO₃

4. aluminum (Al³⁺) carbonate (CO₃²⁻)  - Al₂(CO₃)₃

Learn more about:

formulas for the ionic compounds

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7 0
3 years ago
A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 0
3 years ago
What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?
Natalija [7]

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH              1 mole
--------------------------  x  -----------------------  =  0.355 moles KOH
                                     56.014 grams

(Step 2)

Molarity = moles / volume                            <----- Molarity ratio

0.550 M = 0.355 moles / volume                 <----- Insert values

(0.550 M) x volume = 0.355 moles              <----- Multiply both sides by volume

volume = 0.645 L                                          <----- Divide both sides by 0.550

6 0
1 year ago
Over the last several decades, scientists have addressed the problem of nonrenewable natural resources such as fossil fuels. Hum
Varvara68 [4.7K]

The answer is D

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6 0
3 years ago
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