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2 years ago

According to the Kinetic Molecular Theory, the absolute temperature of a gas is directly related to average molecular kinetic

Chemistry

2 answers:

astra-53 [7]2 years ago

7 0

Answer;

-energy

According to the Kinetic Molecular Theory, the absolute temperature of a gas is directly related to average molecular kinetic energy.

Explanation;

Kinetic molecular theory is a theory that explains that the particles of a gas are in constant random motion, exhibiting perfectly elastic collisions.
This theory may be explained using Boyle's and Charles's law. Such that the average kinetic energy of a given particles of a gas is directly proportion to the absolute temperature.

Alekssandra [29.7K]2 years ago

6 0

Answer: The correct answer is Option D.

Explanation:

Average kinetic energy is defined as the average of kinetic energies of all the particles in a system. It is a measure of temperature. The chemical equation used to calculate average kinetic energy is given as:

$K=\frac{3RT}{2N_A}$

where,

K = average kinetic energy

R = Gas constant

$N_A$ = Avogadro's number

T = temperature of the system

Average kinetic energy of the system is directly proportional to the temperature of the system. Thus, if temperature increases, the average kinetic energy also increases and vice-versa.

From the above information, it is concluded that the correct answer is Option D.

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Copper(l) sulfide can react with oxygen to produce copper metal by the reaction cu2s + o2 => 2cu + so2. if 5.00 g of cu2s is

valentina_108 [34]

To answer this problem, we write first the stoichiometric equation. Thus we have,

Cu2S + O2 => 2Cu + SO2

Next, we check if the equation is balanced or not.

(left) (right)
2 Cu = 2 Cu
1 S = 1 S
2 O = 2 O

So the stoichiometric equation is balanced, let's proceed in solving the theoretical yield of Cu given 5 g of Cu2S.

First, we solve for Cu2S in moles,
5 g Cu2S x 1 mol Cu2S = 0.0314 mol Cu2S
159.16 g Cu2s

Secondly, convert moles of Cu2S to moles Cu. Note for every mole of Cu2S we get 2 moles of Cu. Thus,

0.0314 mo Cu2S x 2 mol Cu = 0.0628 mole Cu
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Lastly, we convert mole Cu to g Cu via multiplying by Cu's MW.

0.0628 mole Cu x 63.546 g Cu = 3.99 g Cu or 4 g Cu
 1 mole Cu

ANSWER: 4 g Cu

4 0

3 years ago

Read 2 more answers

For the reactions below, describe the reactor system and conditions you suggest to maximize the selectivity to make the desired

alexandr1967 [171]

Answer:

hello your question lacks the required reaction pairs below are the missing pairs

Reaction system 1 :

A + B ⇒ D $-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}$

A + B ⇒ U $-r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}$

Reaction system 2

A + B ⇒ D $-r_{1A} = 10exp( -1000K/T)C_{A}C_{B}$

B + D ⇒ U $-r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}$

Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

reaction 2 :

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

Explanation:

Reaction system 1 :

A + B ⇒ D $-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}$

A + B ⇒ U $-r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}$

the selectivity of D is represented using the relationship below

$S_{DU} = \frac{-r1A}{-r2A}$

hence SDu = 1/10 * $\frac{exp(-800K/T)}{exp(-1000K/T)} * C_{A} ^{0.5} C_{B} ^{-0.5}$

description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

Reaction system 2

A + B ⇒ D $-r_{1A} = 10exp( -1000K/T)C_{A}C_{B}$

B + D ⇒ U $-r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}$

selectivity of D

$S_{DU} = \frac{-r1A}{-r2A}$

hence Sdu = $1/10^7 * \frac{exp(-1000K/T)}{exp(-10000K/T)} *\frac{C_{A} }{C_{D} }$

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

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