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4 years ago

Which set of numbers gives the correct possible values of l for n = 2

2 answers:

3 0

HEY THERE!!

For a given principal quantum number or n, the corresponding angular quantum number or l is equivalent to a range between 0 and (n-1) .

This means that the angular quantum number for a principal quantum number of 2 is equivalent to:

l = 0 -> (n-1) = 0 -> (2-1) = 0 -> 1

So the answer is 0, 1

HOPE IT HELPED YOU.

lina2011 [118]4 years ago

3 0

Answer:

The possible values are l={0,1}

Explanation:

Hi,the possible values that the Azimuthal number (l) are given by the value of the Principal quantum number (n) of the atomic level you are in:

$l=[0;n-1]$

In this case:

For $n=2 \longrightarrow n-1=2-1=1$ so l values will be 0 and 1

This different values the Azimuthal numbar can have, determine the different types of sublevels that atomic level has.

For example:

l=0 is sublevel s
l=1 is sublevel p
l=2 is sublevel d

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3 years ago

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Use the periodic table to calculate the molar mass of each of the following compounds. Give your answer to the correct number of

Musya8 [376]

1) Answer is: molar mas of ammonia is 17.031 g/mol.

M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.

M(NH₃) = 14.007 + 3 · 1.008 · g/mol.

M(NH₃) = 17.031 g/mol.

2) Answer is: molar mas of lead(II) chloride is 278.106 g/mol.

M(PbCl₂) = Ar(Pb) + 2 · Ar(Cl) · g/mol.

M(PbCl₂) = 207.2 + 2 · 35.453 · g/mol.

M(PbCl₂) = 278.106 g/mol.

3) Answer is: molar mas of acetic acid is 60.052 g/mol.

M(CH₃COOH) = 2 · Ar(C) + 2 · Ar(O) + 4 · Ar(H) · g/mol.

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4 years ago

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One mole of carbon (12.0 g) in the form of crystalline graphite is burned at 25◦C and 1.000 atm pressure to form CO2(g). All of

iogann1982 [59]

Answer:

T₂ = 43.46 °C

Explanation:

Given that:

The heat of the formation of carbon dioxide = - 393.5 kJ/mol (Negative sign suggests heat loss)

It means that energy released when 1 mole of carbon undergoes combustion = 393.5 kJ = 393500 J

Heat gain by water = Heat lost by the reaction

Thus,

$m_{water}\times C_{water}\times \Delta T=Q$

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Mass of water = 5100 g

Specific heat of water = 4.18 J/g°C

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3 years ago

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol

spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc. 0.10 0.10 0.50

At eqm. (0.10-x) (0.10-x) (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713 and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

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Thus, the concentration of HI (g) at equilibrium is, 0.643 M

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