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nikitadnepr [17]
3 years ago
8

 A reaction takes place between two gases at 298 K. There are 2.0 moles of gas A and 3.0 moles of gas B. Which scenario would lo

wer the collision frequency between gases as well as the rate of reaction between gases A and B?A)Increase the moles of gas B.B)Lower the temperature of the reaction to 250 K.C)Compress the gases into a smaller volume.D)Introduce a catalyst to the container.
Chemistry
2 answers:
Free_Kalibri [48]3 years ago
7 0

Answer:

B)Lower the temperature of the reaction to 250 K.

Explanation:

Option B is correct. Lowering the temperature will decrease the kinetic energy of the gas molecules decreasing the frequency of collision and the rate of the reaction.

Option A is incorrect. Increasing the moles or the number of gas molecules of A and B will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option C is incorrect. Compressing the gas into a small volume will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option D is incorrect. A catalyst works by lowering the activation energy and thus increasing the frequency and orientation of the collision between gas molecules and thus increasing the rate of the reaction.  

Soloha48 [4]3 years ago
3 0
Choice B I think, because if the temperature decreases the particles move slower due to less kinetic energy, and thus won't collide as frequently.
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How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below?
natka813 [3]

Answer:

8.1433 g of XeF₆  are required.

Explanation:

Balanced chemical equation;

XeF₆ (s) + 3H₂ (g)   →  Xe (g) + 6HF (g)

Given data:

Volume of hydrogen = 0.579 L

Pressure = 4.46 atm

Temperature = 45 °C (45+273= 318 k)

Solution:

First of all we will calculate the moles of hydrogen

PV = nRT

n = PV/ RT

n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K

n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹

n = 0.0995 mol

Mass of hydrogen:

Mass = moles × molar mass

Mass =  0.0995 mol × 2.016 g/mol

Mass =  0.2006 g

Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

                                         H₂   :  XeF₆

                                          3    :    1

                                 0.0995   : 1/3× 0.0995 = 0.0332 mol

Now we will calculate the mass of XeF₆.

Mass = moles × molar mass

Mass = 0.0332 mol × 245.28 g/mol

Mass = 8.1433 g

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Answer:

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Explanation:

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A chemical property of isopropanol : D. Isopropanol is flammable.

<h3>Further explanation </h3>

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Physical changes do not form new substances, so the properties of the particles remain the same.(size,volume,shape)

Example : boiling and freezing, just change its phase form from liquid to gas or from liquid to solid

 

Chemical changes/reaction form new substances(products) that are different from the initial substances(reactants)

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