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3 years ago

How many unpaired electrons are present in the ground state Carbon atom?

Chemistry

1 answer:

blsea [12.9K]3 years ago

8 0

Answer:

Your Answer would be 2 unpaired electrons.

Explanation:

There are 2 unpaired electrons present in the ground state of Carbon Atom.

Hope it helps!

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Suppose of ammonium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of ammonium c

Marta_Voda [28]

Answer:

Final molarity of ammonium cation in the solution = 0.16 M

Explanation:

Complete Question

Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution

2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃

We first convert the given parameters to number of moles

Number of moles = (Mass/Molar mass)

Molar mass of NH₄NO₃ = 80.043 g/mol

Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole

Number of moles = (Concentration in mol/L) × (Volume in L)

Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole

2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄

So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.

So, the number of moles of ammonium ion in the product is also 0.03224 mole.

Molarity = (Number of moles)/(Volume L)

Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M

Hope this Helps!!!

3 0

3 years ago

Utilizing dimensional analysis find the number of inches in a kilometer given that 1 inch equals 2.54 cm

hodyreva [135]

To solve this problem, separate it into chunks that you know. You know that there are 2.54 centimeters in 1 inch. You know that there are 100 centimeters in 1 meter. You know that there are 1000 meters in a kilometer. Therefore, we'll convert in this order: 1) from kilometers to meters, 2) from meters to centimeters, and 3) from centimeters to inches.
1) 1km × 1000m/1km
= 1000m
2) 1000m × 100cm/1m
= 100000cm
3) 100000cm × 1in/2.54cm
≈ 39370in
So, there are approximately 39370 inches in a kilometer.

4 0

4 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

Balance the following chemical equation Na + Cl2 -> NaCl and explain how the balanced equation models the law of conservation

Alexus [3.1K]

Answer:

2 Na + 1 Cl2 -> 2 NaCl

Explanation:

The answer is really simple, because if you have 1 nonmetal element that has a subscript of 2, you need to multiply the product and the first reactant by 2 to balance it.

7 0

3 years ago

U-236 spontaneously decays to Br-87, X and three neutrons. What is element “X”? 1. Th 2. Pb 3. Ra 4. La 5. Ba

Elena-2011 [213]

<span>U-236 spontaneously decays to Br-87, X and three neutrons. The element X is 4. La, also known as Lanthanum, number 57 in the periodic system of elements.</span>

6 0

3 years ago

How many unpaired electrons are present in the ground state Carbon atom?​

How many unpaired electrons are present in the ground state Carbon atom?