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3 years ago

What is the second quantum number of the 3p1 electron in aluminum 1s22s22p63s23p1?

2 answers:

5 0

There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital

here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.

DIA [1.3K]3 years ago

4 0

Answer: I = 1 for apex

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Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica

kiruha [24]

Answer : The Lewis-dot structure of $NH_3$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $NH_3$

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in $NH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on N}=5-2-\frac{6}{2}=0$

$\text{Formal charge on }H_1=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_2=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_3=1-0-\frac{2}{2}=0$

Hence, the Lewis-dot structure of $NH_3$ is shown below.

3 0

3 years ago

45 Three samples of the same solution are tested, each with a different indicator. All three indicators, bromthymol blue, bromcr

ANEK [815]

Answer: The correct answer is Option 4.

Explanation:

Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.

The pH range and color change of these indicators are:

Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6

As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.

Hence, the correct answer is Option 4.

3 0

3 years ago

Read 2 more answers

Can someone help please? :)

Leno4ka [110]

1) This is a definition.

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7 0

2 years ago

Can anyone solve this?

Sonja [21]

Answer:

3P2O5:

P:6 O:15

5CO2

C:5 O:10

3C6H12O6

C:18 H:36 O:18

4C6H12

C:24 H:48

3Mg3(PO4)2

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4 0

3 years ago

At 25 °C, how many dissociated OH– ions are there in 1243 mL of an aqueous solution whose pH is 2.07?

coldgirl [10]

Answer: The number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

Explanation:

We are given:

pH = 2.07

Calculating the value of pOH by using equation, we get:

$2.07+pOH=14\\\\pOH=14-2.07=11.93$

To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:

$pOH=-\log[OH^-]$

We are given:

pOH = 11.93

Putting values in above equation, we get:

$11.93=-\log[OH^-]$

$[OH^-]=10^{-11.93}=1.17\times 10^{-12}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = $1.17\times 10^{-12}M$

Volume of solution = 1243 mL = 1.243 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, $1.424\times 10^{-12}mol$ number of $OH^-$ will contain = $(1.424\times 10^{-12}\times 6.022\times 10^{23})=8.57\times 10^{11}$ number of ions

Hence, the number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

3 0

3 years ago