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3 years ago

Diffrence btwn a period and a group

1 answer:

8 0

Rows of elements are called periods. The period number of an element signifies the highest unexcited energy level for an electron in that element.

Columns of elements help define element groups. Elements within a group share several common properties. Groups are elements have the same outer electron arrangement.

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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

4 years ago

PLEASE HELP I WILL GIVE BRAINLIEST!!! Its due tonight at 12:00

Dmitry [639]

Answer:

Explanation:

we have only distance vs time graph so c is the right answer that cart is moving with constant speed of 0.5m/s

confirmed

first calculate distance which is from 2 to 5.5

5.5-2=3.5

apply the formula of velocity

v=s/t

v=3.5/7=0.5

8 0

3 years ago

Read 2 more answers

How many moles are in 24.75 g of H2O? What is given the conversation and the unknown?

riadik2000 [5.3K]

Answer:

1.373 mol H₂O

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

24.75 g H₂O

Step 2: Identify Conversions

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

$24.75 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O} )$ = 1.37347 mol H₂O

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

1.37347 mol H₂O ≈ 1.373 mol H₂O

7 0

3 years ago

When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, what hybrid orbitals are used by p

SSSSS [86.1K]

When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, hybrid orbitals that are used by phosphorus in the pcl4 cations are one orbital of s and three orbital of p as it is sp³hydridised.

<h3>What is sp³ hybridization?</h3>

Hybridization is a process or system which specifies the shape and geometry of any element or molecule with bond angles too.

The pcl4 cation is sp³ hybridized because of the phosphorus behave as a central atom here and the 4 chloride molecules are attached with the p- orbitals to the phosphorus molecule.

Therefore, pcl4 cation is sp³ hybridized.

Learn more about sp³ hybridization, here :

brainly.com/question/13062274

#SPJ4

5 0

2 years ago

9. the answer is not 100.0(ect.), plese help me

ArbitrLikvidat [17]

2.56 is the answer !!!!

5 0

3 years ago

Read 2 more answers