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Travka [436]
4 years ago
14

when a metal, such as lead, is oxidied (loses electrons) to form a positive ion (cation), how does he solubility change?

Engineering
1 answer:
o-na [289]4 years ago
8 0

Answer: The size of the ion and the charge of the ion are the factors that affect solubility in water.

Explanation:

Lead lose electrons to become cations. Compounds with small ions tend to be less soluble than compounds with large ions. Large ions have higher solubility. This is because small ions are closely packed so it is difficult for water to break them apart.

Compounds with small ions seemingly have less solubility than those with large ions. The ions in the compound attract each other, and the water molecules attract the ions. Compounds would be soluble in water If the water molecules have a greater or higher attraction to the ions than ions have for each other.

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An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

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Calculate the current in the 8-W resistor of Figure below by using Thevenin’s theorem. What will be its value of connections of
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3 years ago
Problem 4 You are designing a circuit to drive LED1, using the following circuit. The datasheet for the LED specifies that VF =
HACTEHA [7]

Answer:

a) I_LED= 1/6 A  b) Vf= 2.5V

Explanation:

Consider circuit in the attachment.

a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)

Solving the circuit in attachment 2 using mesh analysis

-9+2I1+4(I1-I2)-4+2I1=0

8I1 - 4I2= 13 ............... eq 1

4+4(I2-I1)+ I2 + 2=0

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I1= 41/24 ;  I2 = 1/6; I2= I_LED

b) Solving the circuit in attachment 2 again, this time I2=0

8I1 - 4I2= 13

8I1- 4(0)=13

I1= 13/8

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I2=I_LED=0

Vf= 2.5 V

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3 years ago
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