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Inessa05 [86]
2 years ago
10

Which of the following must be the same before and after a chemical reaction?

Physics
1 answer:
madam [21]2 years ago
6 0

Answer: Option (D) is the correct answer.

Explanation:

A chemical reaction is defined as the reaction which causes change in chemical composition of a substance.

According to law of conservation of mass, mass can neither be created nor it can be destroyed. It can only be changed from one form to another.

For example, H^{+} + Cl^{-} \rightarrow HCl

Here, total mass of reactants = (1.008 + 35.453) g/mol = 36.461 g/mol

Whereas total mass of products = 36.461 g/mol

This shows that mass remains the same before or after the reaction.

Similarly, the number of atoms of the type involved will remains the same as no change in their nucleus is occurring.

Thus, we can conclude that the following must be the same before and after a chemical reaction.

  • The sum of the masses of all substances involved.
  • The number of atoms of the type involved.
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Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Mirrors exhibit specular reflection.

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An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
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Answer:

1) A

2) C

3) B

4) A

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6) Incomplete information(picture missing)

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8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

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6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

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