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2 years ago

What pressure is required to compress 196.0 liters of air at 1.83 atmosphere into a cylinder whose volume is 26.0 liters

Chemistry

1 answer:

drek231 [11]2 years ago

5 0

Answer:

P₂ = 13.79 atm

Explanation:

Given data:

Initial volume = 196.0 L

Initial pressure = 1.83 atm

Final volume = 26.0 L

Final pressure = ?

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.83 atm × 196.0 L = P₂× 26.0 L

P₂ = 358.68 atm. L / 26.0 L

P₂ = 13.79 atm

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Answer:

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Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

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Which situation would cause the following equilibrium reaction to decrease the formation of the products? 2SO2 (g) + O2 (g) Two

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The given equilibrium reaction is,

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The given reaction is exothermic. So, heat energy will be a product. Therefore, decreasing the temperature (heat energy) would lead to the formation of more products as when the amount of energy which is a product is reduced, there is more room for the products to form.

Increasing the pressure would shift the equilibrium towards that side which has least number of moles of the gaseous substance. Hence, here increasing the pressure would lead to the formation of more products by shifting the equilibrium towards the right side.

Decreasing the volume would make the equilibrium shift towards the least number of moles of the gaseous substance. So, here in this equilibrium decreasing the volume would lead to the formation of more products.

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tatiyna

Answer:

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Explanation:

Step 1: Given data

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Mass of the filled container (m₂): 105.22 g

Volume of the unknown liquid (V): 20.14 mL

Step 2: Calculate the mass of the liquid

The mass of the liquid is equal to the difference between the mass of the filled container and the mass of the empty container.

$m = m_2 - m_1 = 105.22g - 80.21 g = 25.01 g$

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