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ExtremeBDS [4]
3 years ago
8

Which sequence represents the relationship between pressure and volume of an ideal gas as explained by the kinetic-molecular the

ory?
A. more gas particles - more collisions - higher pressure

B. smaller volume - crowded particles - less collisions - lower pressure

C. smaller volume - crowded particles - more collisions - higher pressure

D. more gas particles - more kinetic energy - more volume - higher pressure
Chemistry
2 answers:
Svetlanka [38]3 years ago
8 0
<h2>Answer : Option C) Smaller volume - crowded particles - more collisions - high pressure</h2><h3>Explanation : </h3>

The kinetic molecular theory of gases explains that if there is small volume of gas there will be more crowding of the gas molecules inside the container. The crowded gas molecules will collide with each other and also with the walls of container as a result, exchange of energies will take place. Which will increase the pressure inside the container, and will raise the pressure than the initial pressure.

Dafna11 [192]3 years ago
6 0
The correct answer is C i just took the test on e2020
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How much carbon dioxide will be formed if 12.5 grams of oxygen reacts with 7.2 grams of propane (C3H8 )? Balanced equation: C3H8
solong [7]

Answer:

10.3125 grams of carbon dioxide will be formed.

Explanation:

The balanced reaction is:

C₃H₈ + 5 O₂→ 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of the compounds:

  • C₃H₈: 44 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

then by stoichiometry of the reaction, the following amounts of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole = 44 g
  • O₂: 5 moles* 32 g/mole= 160 g
  • CO₂: 3 moles* 44 g/mole= 132 g
  • H₂O: 4 moles* 18 g/mole= 72 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: If by reaction stoichiometry 44 grams of propane react with 160 grams of oxygen, 7.2 grams of propane react with how much mass of oxygen?

mass of oxygen=\frac{7.2 grams of propane*160 grams of oxygen}{44 grams of propane}

mass of oxygen= 26.18 grams

But 26.18 moles of O₂ are not available, 12.5 grams are available. Since you have less mass than you need to react with 7.2 grams of propane, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry of the reaction 160 grams of oxygen form 132 grams of carbon dioxide, 12.5 grams of oxygen will form how much mass of carbon dioxide?

mass of carbon dioxide=\frac{12.5 grams of oxygen*132 grams of carbon dioxide}{160 grams of oxygen}

mass of carbon dioxide= 10.3125 grams

<u><em>10.3125 grams of carbon dioxide will be formed.</em></u>

<u><em> </em></u>

<u><em></em></u>

7 0
3 years ago
Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di
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Answer : The correct rate law for the reaction is,

\text{Rate}=k[CHCl_3][Cl_2]^{1/2}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)

Rate law expression for the reaction:

\text{Rate}=k[CHCl_3]^a[Cl_2]^b

where,

a = order with respect to CHCl_3

b = order with respect to Cl_2

Expression for rate law for first observation:

0.0035=k(0.010)^a(0.010)^b ....(1)

Expression for rate law for second observation:

0.0069=k(0.020)^a(0.010)^b ....(2)

Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

0.027=k(0.040)^a(0.040)^b ....(4)

Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

Dividing 2 from 3, we get:

\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}

Calculation used :

1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}

Thus, the rate law becomes:

\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}

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H2O has hydrogen bonding, which is a form of dipole-dipole forces, and NO3- is an ion, so the intermolecular attraction is ion-dipole.
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