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Zina [86]
3 years ago
11

Consider the reaction: 2 SO2(g) + O2(g) <----> 2 SO3. If, at equilibrium at a certain temperature, [SO2] = 1.50 M, [O2] =

0.120 M, and [SO3] = 1.25 M, what is the value of the equilibrium constant? none of the above 8.68 6.94 0.14 5.79
Chemistry
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

5.79

Explanation:

The following data were obtained from the question:

Concentration of SO2, [SO2] = 1.50 M

Concentration of O2, [O2] = 0.120 M

Concentration of SO3, [SO3] = 1.25 M

Equilibrium constant, (K) =..?

The balanced equation for the reaction is given below:

2SO2(g) + O2(g) <==> 2SO3(g)

From the balanced equation, the equilibrium constant K is written as follow:

K = [SO3]²/ [SO2]²• [O2]

With the above, the value of the equilibrium constant, K can be obtained as follow

K = [SO3]²/ [SO2]²• [O2]

K = (1.25)² / (1.50)² × 0.120

K = 1.5625 / ( 2.25 x 0.120)

K = 5.79

Therefore, the equilibrium constant is 5.79.

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A species that is isoelectronic with the nitrate ion and hence would have the same shape is [1] [2] [3] [4] [5] sulphur trioxide
Bumek [7]

Answer:

[1] sulphur trioxide

Explanation:

Isoelectronic species have the same number of valence electrons.

Valence electrons in nitrate (NO₃⁻):

5e- (N) + (3 x 6e-)(3xO) + 1e- (charge) = 24e-

Valence electrons in sulphur trioxide (SO₃):

6e- (S) + (3 x 6e-)(3xO) = 24e-

Valence electrons in sulphite (SO₃²⁻):

6e- (S) + (3 x 6e-)(3xO) + 2e- (charge) = 26e-

Valence electrons in phosphine (PH₃):

5e- (P) + (3 x 1e-)(3xH) = 8e-

Valence electrons in water (H₂O):

6e- (O) + (2 x 1e-)(2xH) = 8e-

Valence electrons in chlorite (ClO₂⁻):

7e- (Cl⁻) + (2 x 6e-)(2xO) + 1e- (charge) = 20e-

The only species isoelectronic with nitrate is sulphur trioxide. Both have trigonal planar geoemetry.

8 0
3 years ago
In a molecule of chlorine (Cl2), the chlorine atoms are connected to each other by a single bond. Which statements correctly des
blsea [12.9K]

Answer:

It is a sigma bond

Explanation:

Chlorine has an electronic configuration of 1s2 2s2 2p6 3s2 3p5. This means that the outermost n=3 level has seven electrons. Hence one more electron is needed for the octet of outermost electrons to be achieved. As a result of this, chlorine enters into covalent bonding with another chlorine atom to form Cl2.

The outermost 3p electrons of the two chlorine atoms are now shared to form a p-p sigma bond (a single bond). Hence, the Cl2 molecule contains a sigma(single) bond between two chlorine atoms. Hence the answer written above.

7 0
3 years ago
If my primary DNA strand is ATACCGCAA <br> a write the complimentary DNA strand
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Answer:

TATGGCGTT

Explanation:

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A-T

C-G

Use the other letter for complimentary strands

6 0
3 years ago
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Which equation shows how to calculate how many grams (g) of KOH would be needed to fully react with 4 mol Mg(OH)2? The balanced
bazaltina [42]
Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)

moles KOH = 4 x 2 / 1

= 8 moles of  KOH

molar mass KOH = 56 g/mol

mass of KOH = n x mm

mass of  KOH = 8 x 56

= 448 g of KOH 

hope this helps!

3 0
3 years ago
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Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

7 0
3 years ago
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