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Naya [18.7K]
2 years ago
12

A student Hatfields a jar with rice and take Insta jars lids the student wants to use a jar of rice to model particle motion of

a substance in any state of matter
Chemistry
1 answer:
Sliva [168]2 years ago
8 0

Answer and Explanation:

The student can use the half rice pot to model the movement of the particles of the material. For this it is necessary that he research the spatial arrangement of the molecules in a solid material, a liquid and a gas.

In this research, he will find that the molecules in a solid material are positioned very close to each other, not allowing any of them to move. In this case, he can pass the grain of hoops to a smaller pot, which is completely filled by the grains. When shuffling the pot, the boy will see that the rice grains do not move and assuming that each grain represents a molecule of the material, he will notice that the molecules in a solid material do not move.

The research will show that the molecules of a liquid material are a little more distant from each other than the molecules of solid materials. However, this distance is not very long, allowing these molecules to be able to touch each other when they are in motion. To visualize this with the rice grains, the student must remove some grains from the jar and allow a certain space between one grain and another.

The molecules of a gaseous material are very far apart. These molecules have access to full movement and rarely touch each other. To visualize this, the student must remove many grains of rice from the pot, leaving it almost empty and watching the movement of the remaining grains.

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3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
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∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

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mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

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(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

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⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

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