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Mazyrski [523]
3 years ago
11

A hollow conductor is positively charged. A small uncharged metal ball is lowered by a silk thread (insulating material) through

a small opening in the top of the conductor and allowed to touch its inner surface. After the ball is removed, it will have: A. A positive charge B. A negative charge C. A charge whose sign depends on where the small hole is located in the conductor D. A charge whose sign depends on what part of the inner surface it touched E. No appreciable charge
Chemistry
1 answer:
laiz [17]3 years ago
4 0

Answer: Option (E) is the correct answer.

Explanation:

Since, the conductor is hollow which means that it is opened on both the ends. Hence, when a small uncharged metal ball is passed through it with the help of a silk thread then due to the presence of this insulating thread the ball will not come directly in contact with the charged rod.

As a result, there will occur no formation of opposite charge on the metal ball. Therefore, the ball will remain uncharged in nature.

Thus, we can conclude that after the given ball is removed, it will have no appreciable charge.

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What is the concentration of chloride in a solution made with 0.808 grams of CaCl2 and 250.0 ml of water.?
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Answer:

0.0584 M

Explanation:

From the question given above, the following data were obtained:

Mass of CaCl₂ = 0.808 g

Volume of water = 250 mL

Concentration of chloride =?

Next, we shall determine the number of mole in 0.808 g of CaCl₂. This can be obtained as follow:

Mass of CaCl₂ = 0.808 g

Molar mass of CaCl₂ = 40 + (35.5 × 2)

= 40 + 71

= 111 g/mol

Mole of CaCl₂ =?

Mole = mass / Molar mass

Mole of CaCl₂ = 0.808 / 111

Mole of CaCl₂ = 0.0073 mole

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Next, we shall determine the molarity of CaCl₂. This can be obtained as follow:

Mole of CaCl₂ = 0.0073 mole

Volume of water = 0.25 L

Molarity of CaCl₂ =?

Molarity = mole /Volume

Molarity of CaCl₂ = 0.0073 / 0.25

Molarity of CaCl₂ = 0.0292 M

Finally, we shall determine the concentration of the chloride as illustrated below:

CaCl₂ <=> Ca²⁺ + 2Cl¯

From the equation above,

1 mole of CaCl₂ produced 2 mole of Cl¯.

Therefore, 0.0292 M CaCl₂ will produce = 0.0292 × 2 = 0.0584 M Cl¯.

Thus, the concentration of the chloride ion (Cl¯) in the solution is 0.0584 M

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