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4 years ago

Which will not appear in the equilibrium constant expression for the reaction below?

Chemistry

1 answer:

n200080 [17]4 years ago

5 0

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

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How much iodine (I2), in grams, should be added to water to produce 2.5L of solution with a molarity of 0.56M?

denis-greek [22]

Molarity=Moles of solute/Volume of solution in L

0.56M=moles/2.5L
moles=0.56(2.5)
moles of Iodine=1.4mol

Mads of Iodine

Moles(Molar mass)
1.4(126.9)
177.66g

7 0

2 years ago

Given the formula for an organic compound:

Dmitriy789 [7]

Based on the picture,
The name given to the group in the box would be : Ethyl

You can see that in the picture it has 2 carbon side chains, which mean it categorized as Ethyl

hope this helps

3 0

3 years ago

Read 2 more answers

An athlete preparing for a marathon runs 21.3 miles. How many kilometers did they run?

densk [106]

Answer: 34.27903 km

Explanation: for an approximate result, multiply the length value by 1.609 to 21.3 to get answer.

8 0

3 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

3 years ago

Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to

statuscvo [17]

Answer:

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

Explanation:

From the question it is clear that,

Initial volume of sodium sulphide solution is (v₁) = 50mL

Initial concentration of sodium sulphide solution is (s₁) =0.874 M

Final volume of sodium sulphide solution is (v₂) = 250mL

Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,

v₁ × s₁ = v₂ × s₂

Or, s₂ = v₁ × s₁/v₂

= 50 × 0.874 / 250

= 0.1748 M

Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

5 0

3 years ago