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klemol [59]
3 years ago
5

What percentage of carbon's orbitals are spherical in conformation?

Chemistry
2 answers:
Andrej [43]3 years ago
8 0

Answer:

40 percent.

Explanation:

In order to be able to solve this question effectively one has to understand quantum Chemistry very well, so I will recommend that you read more on quantum chemistry.

Schrodinger did a lot of work on orbitals, probabilities of finding an electron on orbitals and how electrons are distributed.

So, instead of taking the complex way to solve this question, let us take the simplest path or way to solve the question. Given the electronic structure below;

1s2 2s2 2px1 2py1 2pz0.

That is, carbon contains 5 orbitals, that is two(2) s- orbitals and three(3) p- orbitals. Recall that s-orbitals is spherical in shape, then 2/5 × 100 = 40%.

Margarita [4]3 years ago
3 0
Orbitals are space around the nucleus where electrons orbit. Several orbitals can exist in a given energy level/shell. Carbon is the sixth element of the periodic table with 6 electrons. Carbon atoms have the ability to bond to themselves and to other atoms with sp, sp² and sp³, hybrid orbitals. Several orbitals in carbon are spherical conformation and this accounts to about 40% of the orbitals.
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What is the maximum mass of ammonia that can be formed when 36.52 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas
k0ka [10]

The maximum mass of NH₃ that can be formed when 36.52 g of N₂ reacts with 10.62 g of H₂ is 44.35 g

<h3>Balanced equation </h3>

N₂ + 3H₂ —> 2NH₃

Molar mass of N₂ = 14 × 2 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 3 × 2 = 6 g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

SUMMARY

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂ to produce 34 g of NH₃

<h3>How to determine the limiting reactant </h3>

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

36.52 g of N₂ will react with = (36.52 × 6) / 28 = 7.83 g of H₂

From the above calculation, we can see that only 7.83 g out of 10.62 g of H₂ are required to react completely with 36.52 g of N₂.

Therefore, N₂ is the limiting reactant

<h3>How to determine the maximum mass of NH₃ produced </h3>

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃

Therefore,

36.52 g of N₂ will react to produce = (36.52 × 34) / 28 = 44.35 g of NH₃

Thus, the maximum mass of NH₃ obtained from the reaction is 44.35 g

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

4 0
2 years ago
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

4 0
3 years ago
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