Ask question

Ask question

All categories

2 years ago

5

Revise the following statement to make it true:

1 answer:

tino4ka555 [31]2 years ago

6 0

Answer:

All the individuals of a species living within a specific area are collectively called a population.

Explanation:

You might be interested in

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

Which of these is an example of heat transferred by radiation?

timofeeve [1]

A I think I’m not sure

5 0

3 years ago

Read 2 more answers

A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz

Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

$\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m$

Using Rydberg's Formula:

$\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where:

R is Rydberg constant= $1.097*10^7$

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

$\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99$

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

3 0

3 years ago

Briefly describe the solvation of sodium chloride to form aqueous solution.

charle [14.2K]

Answer:

idk

Explanation:

sorry

7 0

3 years ago

Read 2 more answers

8.38e-21 Q^2 -1.07e-23 Q^2 +3.15e+19<br> Solve for Q

GarryVolchara [31]

8.38e -21Q^2 -1.07e -23Q^2 +3.15e +19
= 10.46e -44Q^2 + 19
44Q^2 = 10.46e + 19
Q^2 = 523/2200e + 19/44
Q1= ≈ -1.03828
Q2= ≈ 1.03828

4 0

3 years ago