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Bond [772]
3 years ago
8

The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's char

ge?
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

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Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of
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Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

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\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles

\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2P+5Cl_2\rightarrow 2PCl_5

From the balanced reaction we conclude that

As, 5 moles of Cl_2 react with 2 moles of P

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of P

That means, in the given balanced reaction, Cl_2 is a limiting reagent and it limits the formation of products and P is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of PCl_5.

As, 5 moles of Cl_2 react with 2 moles of PCl_5

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of PCl_5

Now we have to calculate the mass of PCl_5.

\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5

\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g

Now we have to calculate the mass of product produced (actual yield).

\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100

70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100

\text{Actual yield of }PCl_5=20.67g

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

3 0
3 years ago
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