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3 years ago

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A solution is formed by dissolving 83.2 grams of copper II chloride (CuCl2) in 2.5 liters of water. The molar mass of CuCl2 is 1

34.45 g/mol.
What is the molarity of the solution?

1 answer:

alexandr1967 [171]3 years ago

6 0

Explanation:

the answer is in the above image

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When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory, what hybrid orbitals are used by

Vlad1618 [11]

Answer:

sp³

Explanation:

Number of hybrid orbitals = ( V + S - C + A ) / 2

Where

H is the number of hybrid orbitals

V is the valence electrons of the central atom = 5

S is the number of single valency atoms = 4

C is the number of cations = 1

A is the number of anions = 0

For PCl₄⁺

Applying the values, we get:

H = ( 5+4-1+0) / 2

= 4

<u>This corresponds to sp³ hybridization.</u>

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xeze [42]

The correct answers are A and C.

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What is the difference between weight and mass?

Marina86 [1]

Key differences between Mass and Weight

The weight may vary, but the mass is constant.

The mass is measured in kilograms (kg), while the weight is measured in newtons (N).

Mass refers to the amount of matter an object has, but the weight refers to the force of gravity acting on an object.

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If the distance between two objects is decreased to - of the original

Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

$F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R$

And the force at this distance would be written in terms of the same equation:

$F_2 = G\frac{m_1m_2}{R_2^2}$

Find the ratio between the final and the initial force:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}$

Substitute the value for the final distance in terms of the initial distance:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}$

Simplify:

$\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}$

This means the new force will be \frac{1}{100} of the original force.

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U wanna talk with me hit me up

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Answer:

No thanks bro.

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