When a magnet spins within a coil of wires, what is produced?

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

Which of these periods contain elements with electrons in s, p, d, and f orbitals?

zmey [24]

I think the answer is D. Periods 6-7, Because the periods that would contain elements with electrons in s, p, d, and f orbitals are periods 6-7.

(I hope this helps you a lot! :) Have a nice day!)

7 0

3 years ago

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential

nika2105 [10]

Answer:

3.50*10^-11 mol3 dm-9

Explanation:

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.

Ag2C2O4 --> 2Ag+ + C2O4 2-

So Ksp = [Ag+]^2 * [C2O42-]

In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.

So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)

= 3.50*10^-11 mol3 dm-9

7 0

3 years ago

Help me please and only answer if you know what the answer is

avanturin [10]

They have similar chromosomes

3 0

3 years ago

For each mole of glucose (C6H12O6) completely oxidized by cellular respiration, how many moles of CO2 are released in the citric

topjm [15]

Answer:

4

Step-by-step explanation:

The reactions are:

Glycolysis: 1 glucose ⟶ 2 pyruvate

Link reaction: 2 × [1 pyruvate ⟶ 1 acetyl CoA]

Citric acid cycle: 2 × [1 AcetylCoA ⟶ 2 CO₂]

Now, add the reactions, cancelling species that occur on both sides of the reaction arrow,

1 glucose ⟶ 2 pyruvate

2 pyruvate ⟶ 2 acetyl CoA

2 AcetylCoA ⟶ 4 CO₂

Overall : 1 glucose ⟶ 4 CO₂

For each mole of glucose, four molecules of CO₂ are released in the citric acid cycle.

4 0

3 years ago