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Andreas93 [3]
2 years ago
15

Identify the four quantum number for 3p5​

Chemistry
1 answer:
12345 [234]2 years ago
3 0

Answer:

Four quantum numbers of unpaired electron of chlorine are n=3,l=1,m=0,s=+1/2.

Explanation:

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1,100 grams :)
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4 0
3 years ago
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​
11111nata11111 [884]

Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

7 0
3 years ago
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8 0
3 years ago
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mina [271]

Answer:

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5 0
2 years ago
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The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
belka [17]

Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0
3 years ago
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