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3 years ago

11

What is the function of the load in an electric circuit?

2 answers:

mezya [45]3 years ago

8 0

Answer:

The source side of the circuit includes all parts of the circuit between the positive side battery post and the load. The load is any device in the circuit that produces light, heat, sound or electrical movement when current is flowing. A load always has resistance and consumes voltage only when current is flowing.

Explanation:

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Nitella [24]3 years ago

4 0

<h2><u>ANS</u></h2>

<h3>An electric circuit is an energy transformer. It converts energy from one form to another. Electric current is used to drive the load which represents the energy being converted. The final form of the energy may be kinetic, potential, or any other form (entropy, thermal, mechanical, structural, organizational, biological, intellectual, financial, metaphysical, etc.). To achieve this, the electric current does work. They type of work varies depending on the final form of the energy.</h3>

<h2><u>Th</u><u>ank</u> <u>You</u> !!</h2>

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One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

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