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chubhunter [2.5K]
3 years ago
10

1.

Chemistry
1 answer:
Romashka [77]3 years ago
3 0

The voltage needed = 120 V

<h3>Further explanation </h3>

Electric current (symbolized by the letter "i) is the amount of charge flowing in the conductor of each unit of time  

Can be formulated:  

\large{\boxed{\bold{I=\frac{Q}{t}}}

Q = charge (Coulomb)  

t = time (seconds)  

The international standard unit of electric current is Coulomb/second or Ampere (A).  

Electric current is also a ratio between the voltage and the amount of resistance at the conductor.  

Can be formulated  

\large{\boxed{\bold{I=\frac{V}{R}}}

V = Voltage (volts)  

R = resistance (ohms)  

I=current=4 A

R=resistance=30 Ω

The voltage needed :

\tt V=I\times R\\\\V=4\times 30\\\\V=\boxed{\bold{120~V}}}

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What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
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Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

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Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

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0.180 = mole of KOH /0.055

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Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

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From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

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Mole of Fe(OH)3 = 3.3×10¯³ mole

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3.3×10¯³ = Mass of Fe(OH)3 / 107

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Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

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