Got this really tricky extra credit question I can't figure out. Anyone want to give #24 a try?

Radda [10]

Answer:

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

Explanation:

4 0

2 years ago

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

Naddik [55]

Answer:

(i)The mole fractions are :

$A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii) $K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

The initial number of moles of :

$A=1$

$B=2$

$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii)

$K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}$

Where ,

$P_A,P_B,P_C,P_D$ are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

$P_A= 0.0930*1=0.0930$

$P_B= 0.3256*1=0.3256$

$P_C= 0.2093*1=0.2093$

$P_D= 0.3721*1=0.3721$

$K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508$

(iii)

Δ $G=-RTlnK\\$

ΔG = $-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ$

3 0

2 years ago

This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and

n200080 [17]

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

Where [] are concentrations at equilibrium

And Q is:

Q = [HI]² / [H2] [I2]

Where [] are actual concentrations of the reactants.

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

7 0

3 years ago

Which substance, when dissolved in water, forms a solution that conducts an electric current?

Margarita [4]

Missing question:
(1) C2H5OH (3) C12H22O11
(2) C6H12O6 (4) CH3COOH.

Answer is: 4) CH₃COOH, acetic acid.
In water, acetic acid dissociates on ions, so it can conduct electricity:
CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq).
When we put electrodes (cathode and anode) in acetic acid solution, positive and negative ions migrate to electrodes.
Negative acetate ions go to positive electrode and gives electrons to electrode.
Positive hydrogen ions go to negative electrode and gain electrons.

6 0

2 years ago

Solve for x, where M is molar and s is seconds. x=(5.3×103M−2s−1)(0.26M)3

GREYUIT [131]

Answer:

guys can u solve my ques plz

4 0

2 years ago