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sveta [45]
3 years ago
10

Consider the reaction below. C2H4(g) + H2(g) to C2H6(g) Which change would likely cause the greatest increase in the rate of the

reaction?
a decrease temperature and decrease pressure
b increase temperature and decrease pressure
c decrease temperature and increase pressure
d increase temperature and increase pressure
Chemistry
2 answers:
mr_godi [17]3 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

In the given reaction, C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g) greatest increase in rate of reaction only occur when we increase the temperature and also increase the pressure.

This is because increase in pressure will bring the molecules closer to each other whereas increase in temperature will help in more number of collisions between reactant molecules.

Thus, this will lead to increase in formation of products. Hence, we can conclude that rate of reaction will increase with increase in temperature and increase in pressure.

algol [13]3 years ago
4 0
D. Increase temperature and increase pressure
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abruzzese [7]

Answer:

1. CaCO3 + 2HCl —> CaCl2 + H2O + CO2

2. Mg(OH)2 + 2HCl —> MgCl2 + 2H2O

3. Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

7 0
3 years ago
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g)
wolverine [178]

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478.  The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

 AB₂(g)       ↔   <em>A(g)</em> + <em>2B(g)</em>          Kc₁= 1/59

<em>A(g)</em> + <em>3B(g)</em> ↔   AB₃(g)                  Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

6 0
3 years ago
Liquid octane (CH) has a density of 0.7025 g/mL at 20 °C. Find the true mass (murue) of octane when the mass weighed in 18 air i
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Explanation:

According to Buoyance equation,

          m = [m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]

where,      m = true mass

                 m' = mass read from the balance = 17.320 g

              d_{a} = density of air = 0.0012 g/ml

              d_{w} = density of the balance = 7.5 g/ml

                    d = density of liquid octane = 0.7025 g/ml

Now, putting all the given values into the above formula and calculate the true mass as follows.

      m = [m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]    

          = [17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]

          = 17.320 g \times 0.999850                

          = 17.317 g

Thus, we can conclude that the true mass of octane is 17.317 g.

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