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galben [10]
3 years ago
7

What are the two types of collisions? What is the difference between them? Types: a. Types: b. Compare and contrast the two type

s:
Chemistry
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

The two types of collisions are :

Type a)

<u>Elastic collision</u>

Type b)

<u>Inelastic collision</u>

Explanation:

Collision : It is the event when two bodies collide with each other for small period of time.

During collision , the bodies exert force to each other.

Example :

When boxer hits with punches .

When bat hits the ball in cricket match.

So, collision is short duration interaction of two objects. When the objects collides , there is  change in their velocity.

All collision follow law of conservation of momentum . Their type is decided by , whether they follow conservation of energy also.

<u>Compare and contrast the two types</u>

a) Elastic collision : Those collision in which no loss or gain of kinetic energy will occur. They follow conservation of kinetic energy. Example : ideal gaseous molecule

b) Inelastic collision : Those collision in which Change in kinetic energy will occur. They do not follow conservation of kinetic energy.Almost all conservation are inelastic.

Here Kinetic energy get converted into other form of energy.

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Explain how you would decide which represents a greater mass: 3.25 x 1022 atoms of Ca or 2.45 grams of Mg?
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Answer:

  • <u>You need to convert the number of atoms of Ca into mass in grams, using Avogadro's number and the atomic mass of Ca.</u>

Explanation:

The amount of matter is measured in grams. Thus, you need to convert the number of atoms of Ca (calcium) into mass to compare with 2.45 grams of Mg.

To convert the atoms of calcium into mass, you divide by Avogadro's number, to obtain the number of moles of atoms, and then divide by the atomic mass of calcium.

<u />

<u>1. Number of moles, n</u>

      n=\dfrac{\text{number of atoms}}{\text{Avogadro's number}}\\ \\ \\ n=\dfrac{3.25\times 10^{22}}{6.022\times 10^{23}}=0.053969mol

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4 0
3 years ago
Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe
jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

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