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2 years ago

The inertness of the noble gases is due to

Chemistry

1 answer:

jeka57 [31]2 years ago

6 0

Answer:

Explanation:

Noble gases do not have free valence electrons. They have complete octet or dublet Structure

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At STP, what volume of container would be needed to hold 3.2 moles of gas?

Rzqust [24]

78.4 L volume of container is required to hold 3.2 moles of gas.

Explanation:

STP is defined as the standard temperature and pressure of a gas in room temperature conditions. At STP, one mole of the gas which has Avogadro's number of molecules in it will occupy a volume of 22.4 L.
So, one mole of a substance or gas will occupy a volume of 22.4 L then the volume of the container needed for 3.2 moles of gas is calculated by multiplying 22.4 L, standard volume with the moles of the gas 3.2 moles.
Hence, the answer would be 78.4 L.

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3 years ago

Why does the moon appear to be so bright in the sky

Nana76 [90]

Because of the sun reflects onto the moon and makes it brighter.

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2 years ago

Read 2 more answers

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO

elena55 [62]

Answer : The mass of solute in solution is $1.09\times 10^2g$ .

Solution : Given,

Molarity = 0.730 M

Volume of solution = 1.421 L

Molar mass of sodium carbonate = 105.98 g/mole

Formula used for Molarity :

$Molarity=\frac{w}{M\times V}$

where,

w = mass of solute

M = Molar mass of solute

V = volume of solution in liter

Sodium carbonate is solute and water is solvent.

Now put the given values in above formula, we get the mass of solute in solution.

$0.730mole/L=\frac{w}{(105.98g/mole)\times (1.412L)}$

By rearranging the terms, we get

$m=109.247g=1.09\times 10^2g$

Therefore, the mass of solute in solution is $1.09\times 10^2g$ .

4 0

2 years ago

How do chemists explain on a molecular basis

Juli2301 [7.4K]

Explanation:

It happens because particles of gas are in constant random motion. Thus they can collide with the walls of the container causing pressure on the walls.

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2 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

2 years ago