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Pavel [41]
3 years ago
14

An unknown monatomic gas diffuses twice as fast as Br2. What is the identity of this unknown gas?

Chemistry
1 answer:
kirza4 [7]3 years ago
3 0

Answer : The unknown monoatomic gas is, argon.

Explanation :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_2}{R_1})=\sqrt{\frac{M_1}{M_2}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas  = 2R

R_2 = rate of effusion of Br_2 gas  = R

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of Br_2 gas = 160 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{R}{2R})=\sqrt{\frac{M_1}{160g/mole}}

M_1=40g/mol

From the molar mass 40 g/mol we conclude that the unknown monoatomic gas will be, argon.

Therefore, the unknown monoatomic gas is, argon.

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

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3 years ago
How can you raise the average kinetic energy of the water molecules in a glass of water?
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Boil it. Kinetic energy is energy of motion and hot water moleculues move more
8 0
3 years ago
The order of a reaction Z1) is the product of the powers to which the reactant concentrations are raised in the rate law. Z2) ca
irinina [24]

Explanation:

The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant.

The overall order of reaction is the sum of the individual orders of reaction with respect to the reactants.

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6 0
3 years ago
What volume will 0.405 g of krypton gas occupy at STP?
Rufina [12.5K]

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

                  83.8 g of Kr -------------------- 1 mol

                     0.405 g     -------------------  x

                     x = (0.405 x 1) / 83.8

                     x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

                   PV = nRT

- Solve for V

                      V = nRT / P

- Substitution

                      V = (0.0048)(0.082)(273) / 1

- Simplification

                       V = 0.108 / 1

- Result

                       V = 0.108 L

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Which is a characteristic of a synthesis reaction
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Answer:option A
Multiple reactants are used to form one product.
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