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2 years ago

Which represents the correct equilibrium constant expression for the reaction below?

Chemistry

2 answers:

charle [14.2K]2 years ago

6 0

Answer:

the answer is D

Explanation:3

ON EDG

Ad libitum [116K]2 years ago

4 0

Answer:

$\large\boxed{\large\boxed{K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}}}$

Explanation:

1. Chemical equilibrium equation:

$Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}+2Ag(s)$

2. Species

In an equilibrium constant expression you do not include the solid substances; only gases and dissolved substances.

The symbol $(s)$ means solid, thus Cu(s) and Ag(s) shall not appear in your equilibrium constant expression.

The symbol $(aq)$ means in aqueous solution, thus the both $Ag^+$ and $Cu^{2+}$ must appear in the equilibrium constant expression.

3. Equilibrium constant expression.

It is the quotient of the product of the concentrations of the species on the right hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left hand side, each raised to its coresponding coefficient.

$K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}$

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Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

kramer

Answer:

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy is the measure of the degree of disorderness.

In solids, the entropy is very less compared to liquids and gases.

The entropy order is:

solids<liquids<gases

Among the given substances, water in liquid form has a strong intermolecular H-bond.

So, it has also less entropy.

Next acetic acid.

Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.

HCN has slight H-bonding in IT.

Hence, the entropy order is:

Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)

7 0

1 year ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

3 0

2 years ago

Time Remaining: 1:27:31 If 50.0 g of H₂ and 100.0 g of O₂ react, how many moles of H₂O can be produced in the reaction below? 2

Novay_Z [31]

Answer:

Explanation:

2 H₂(g) + O₂(g) → 2 H₂O(g

2 moles 1 mole 2 mole

50 g of H₂ = 50 /2 = 25 moles of H₂

100 g of O₂ = 100 / 32 = 3.125 moles of O₂

So oxygen is the limiting reagent .

3.125 moles of O₂ will react with 6.25 moles of H₂ to give 6.25 moles of H₂O .

Hence moles of H₂O produced = 6.25 moles .

6 0

1 year ago

A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas

Grace [21]

The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL

3 0

2 years ago

Read 2 more answers

A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat

Jet001 [13]

Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.

6 0

2 years ago