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3 years ago

Calculate the number of moles in 24.18 liters of neon gas at STP.

Chemistry

1 answer:

uranmaximum [27]3 years ago

8 0

The number of moles of Neon gas : 1.079

<h3>Further explanation</h3>

Given

24.18 Liters of Neon gas at STP

Required

The number of moles

Solution

There are 2 conditions that are usually used as a reference in chemical calculations (mainly for determining the volume per mole of a gas or the molar volume), STP and RTP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So mol for 24.18 L :

= 24.18 : 22.4

= 1.079 moles

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One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.

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4 years ago

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a) The equilibrium will shift in the right direction.

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$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

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$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

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The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

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