Give one example of an element and one compound.<br><br> Can someone help me with this pls

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

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

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

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

Explanation:

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

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

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

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

b)

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

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:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

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

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

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

3 0

3 years ago

QUESTION 11

Iteru [2.4K]

I actually don’t know but good question

6 0

3 years ago

An obese man was discovered in his air-conditioned hotel room sitting in a chair in front of the television. The air conditioner

Leno4ka [110]

Answer:

It has been approximately 6 hours after death.

Explanation:

This is because between 2-6 hours after death, the body starts becoming stiff from top to bottom, then spreading to the limbs. Since there is only rigor in his upper body, that would mean that with normal temperature and body conditions, it would be 4 or 5 hours after death. But since he is obese and in cold temperature, there is slower progression of rigor, leading to the maximum time in the first rigor mortis phase, 6 hours.

8 0

3 years ago

You're leaving a carnival and you're pumped because a clown just handed you a red, helium- filled balloon. But then, something t

bixtya [17]

Because the helium inside it is lighter than air , hence it rises up
The balloon from our breath contains carbon dioxide which is heavier than air so it doesn’t rise

6 0

3 years ago

Three teaspoons of sucrose(17.1 g) is dissolved in hot water to make a total volume of 250 ml solution. Calculate the molarity

dusya [7]

Molar mass sucrose :

C12H22O11 = 342.29 g/mol

number of moles <span>of the solute :
</span>
n = mass solute / molar mass

n = 17.1 / 342.29

n = 0.05 moles

Volume in liters :

250 mL / 1000 = 0.25 L

therefore :

Molarity = moles of solute / volume

Molarity = 0.05 / 0.25 => 0.2 mol/L

5 0

3 years ago

Give one example of an element and one compound. Can someone help me with this pls