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3 years ago

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You are given a silver rod, copper plate and copper sulphate solution. Explain the arrangements you will do to get coating of co

pper on the silver rod, draw a neat labelled diagram of the above arrangements.

1 answer:

Mandarinka [93]3 years ago

6 0

<u>U</u> <u>VORBELLO</u> <u>FRANÇAIS</u><u>?</u><u>?</u><u>?</u>

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Describe the appearance of the material. Include its color and state (solid or liquid). Pour ¼ cup of water into a small, clear

Alex777 [14]

<u>Answer:</u>

<em>Here the given material is taken and mixed with water.</em>

<u>Explanation:</u>

The amount of material and water taken are same. Hence if it is not soluble in water it should make a dense and flowy paste like material and if it is soluble in water it should this and thicker density of water should remain.

If the amount of water that we are taking is more than the material will float in water if it is not soluble and lighter than water or would sink if it is heavier than water.

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3 years ago

Read 2 more answers

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

3 years ago

Which of the following contribute(s) to most of the mass of an atom?

xenn [34]

Answer:

1.

B) protons and newtrons

2.

D)

3.

C)

4.

A)

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3 years ago

Is aluminum homogeneous or heterogeneous?

inn [45]

<span> aluminum is an element. All elements are pure substances, so that means they are homogenous.
please mark as brainliest

</span>

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3 years ago

Question 6 (1 point)

Morgarella [4.7K]

Answer:

question 6: winter solstice

question 7: rotation is when an object spins around its axis, revolution is when an object travels in a path around another

question 8: the rotational period is equal to the period of revolution for the moon

question 10: true

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3 years ago