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Svetradugi [14.3K]
3 years ago
9

STATES THAT FOR AN ATOM IN THE GROUND STATE, THE NUMBER OF UNPAIRED ELECTRONS IS THE MAXIMUM POSSIBLE AND THESE UNPAIRED ELECTRO

NS HAE THE SAME SPIN
Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

Hund's rule

Explanation:

Hund's rule  states that for an atom in the ground state, the number of unpaired electrons is the maximum possible and these unpaired electrons have the same spin.

I will clearly illustrate this using the ground state of the nitrogen atom. Nitrogen atom contains seven electrons. Its electronic configuration is; 1s2 2s2 2px1 2py1 2pz1.

Notice that the number of unpaired electrons in this configuration is maximum and they are all expected to possess the same spin in accordance with Hund's rule .

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In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form
Mkey [24]
The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein. 

E = Δmc²
where
E is the energy gained or released during the reaction
c is the speed of light equal to 3×10⁸ m/s
Δm is the mass change

(1.715×10³ kJ)(1,000 J/1 kJ) = Δm(3×10⁸ m/s)²
Δm = 1.91×10⁻¹¹ kg
5 0
3 years ago
Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide
Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

            Mass of 1 mole NO₂ = 46 g/mole

            Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0
3 years ago
Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be
mylen [45]

yield = 52.23 %

Explanation:

We have the following chemical reaction:

2 Al (s) + 2 KOH (aq) + 4 H₂SO₄ (aq) + 10 H₂O → 2 KAl(SO₄)₂·12 (H₂O) (s) + 3 H₂ (g)

mass of aluminium = mass of bottle with aluminium pieces - bottle mass

mass of aluminium = 10.8955 - 9.8981 = 0.9974 g

mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if       2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 /  0.03694) × 100

yield = 52.23 %

Learn more about:

reaction yield

brainly.com/question/7786567

#learnwithBrainly

5 0
3 years ago
In 1953, who devopled the model that is shown below
PilotLPTM [1.2K]
What model? can you screenshot it or send a link?
7 0
3 years ago
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
2 years ago
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