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MatroZZZ [7]
2 years ago
8

How many orbitals are there in the third shell (n=3)?

Chemistry
2 answers:
andrezito [222]2 years ago
5 0
The third shell contanis one s orbital,three p orbitals and five d orbitals.Each orbital hold 2 electron so the total number of electron which can be accommodate in third shell are 18 electrons
Dmitrij [34]2 years ago
4 0

Answer:

9

Explanation:

Hello,

In this case, since the third shell of electrons has  the following 3 subshells:

3s, 3p \ and \  3d

- The s subsehll has one orbital for one pair of electrons: s^1 \ and \   s^2.

- The p subsehll has three orbitals for three pairs of electrons: p^1, \ p^2,\ p^3, \ p^4,\ p^5\ and \ p^6\.

- The d subsehll has five orbital for five pairs of electrons d^1, \ d^2,\ d^3, \ d^4,\ d^5,\ d^6,\ d^7,\ d^8,\   d^9\ and \ d^{10}\.

Therefore, the total number of orbitals when n=3 is:

1+3+5=9

Best regards.

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Explanation:

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Lithium iodide has a lattice energy of −7.3×102kJ/mol and a heat of hydration of −793kJ/mol. Find the heat of solution for lithi
Anna007 [38]
Delta H of solution = -Lattice Energy + Hydration 
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>

<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
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Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

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3 years ago
what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH​
Dimas [21]

Answer:

  • The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula  

  • C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =??     v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M     v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

  • C₁ * V₁ = 2 * C₂ * V₂
  • (C₁ * 0.078 L) = (2*  1.25 M * 0.05 L)
  • C₁ = (2*  1.25 M * 0.05 L) / (0.078 L) = 1.6 M  

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

3 0
3 years ago
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