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Aleks04 [339]
3 years ago
14

Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M,

[B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51
Chemistry
1 answer:
Alex787 [66]3 years ago
5 0
Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M, [B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51





Answer . A
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KengaRu [80]

Answer: The density of the object will be 0.573g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of object = 19.6 grams

Volume of object= 34.2cm^3

Putting in the values we get:

Density=\frac{19.6g}{34.2cm^3}=0.573g/cm^3

Thus density of the object will be 0.573g/cm^3

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Which of the following contains The greatest number of representative particles: 1 mole of water molecules, 1 mole of copper ato
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7 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --&gt; 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

6 0
3 years ago
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What would the atomic mass of an atom if its mass is approximately 12 time that of carbon 12​
Stolb23 [73]

Answer:

144 u

Explanation:

The average atomic mass of carbon is approximately 12 u.

Multiply 12 x 12:

12 x 12 = 144

6 0
2 years ago
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