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4 years ago

How many electrons in an atom can share the quantum numbers n = 4 and l = 3?

Chemistry

1 answer:

o-na [289]4 years ago

3 0

Answer:

$\boxed{\text{14}}$

Explanation:

If l = 3, the electrons are in an f subshell.

The number of orbitals with a quantum number l is 2l + 1, so there

are 2×3 + 1 = 7 f orbitals.

Each orbital can hold two electrons, so the f subshell can hold 14 electrons.

$\boxed{\textbf{14 electrons}} \text{ can share the quantum numbers n = 4 and l = 3.}$

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What noble gas configuration does oxygen achieve when bonded?

Greeley [361]

Answer:

1s2 2s2 2p6

Explanation:

ththe noble gas config. of neon

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2 years ago

How many moles do you have in 37.3 g of Co(CrO4)3

shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

37.3 g Co(CrO₄)₃

Step 2: Identify Conversions

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

Step 3: Convert

$37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} )$ = 0.091662 mol Co(CrO₄)₃

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0

3 years ago

Do electrons have psrticle nature or wave nature?

densk [106]

Explanation:

they have wave nature

its correct answer

4 0

3 years ago

A buffer is formed by mixing 100 ml of 0.1 m nh4cl with 50 ml of 0.3 m nh3. to this solution 13 ml of 0.1 m naoh is added. what

Hoochie [10]

I had the same question last week and my answer was C. Try that.

8 0

4 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

4 years ago