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3 years ago

What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g?

Chemistry

1 answer:

BaLLatris [955]3 years ago

7 0

<h2>Hello!</h2>

The answer is:

The third option, 79.8%

To calculate the percent yield we need to divide the actual yield by the theoretical yield, and then, multiply it by 100 in order to find the percent.

So, we from the statement we know the following information:

$ActualYield=220.0g\\TheoreticalYield=275.6g$

Let's use the following formula and substitute the given information:

$PercentYield=\frac{ActualYield}{Theoretical}*100\\\\$

$PercentYield=\frac{220g}{275.6g}*100=0.798*100=79.8(percent)$

Hence, we have that the correct option is the third option, the percent yield is 79.8%.

Have a nice day!

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maks197457 [2]

Answer:

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Explanation:

Data

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3 years ago

_______ is the amount of heat necessary to raise the temperature of 1 gram of a substance by 1 degree Kelvin. A. A calorie B. En

Maksim231197 [3]

<h2>The required option d) "specific heat" is correct.</h2>

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3 years ago

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3 years ago

Read 2 more answers

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3 years ago

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metal weighing 6.98 g was heated to 91.29 °C and then put it into 114.84 mL of water (initially at 24.37 °C). The metal and wate

torisob [31]

Answer:

The specific heat of the metal is 10.93 J/g°C.

Explanation:

Given,

For Metal sample,

mass = 6.98 grams

T = 91.29°C

For Water sample,

volume = 114.84 mL

T = 24.37°C.

Final temperature of mixture = 33.54°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that

water sample temperature changed from 24.37°C to 33.54°C and metal sample temperature changed from 91.29°C to 33.54°C.

We have all values, but, here mass of water is not given. It can be found by using the formula

Density = Mass/Volume

Since, density of water = 1 g/mL

we get, Mass = 114.84 grams.

Since specific heat of water is 4.184 J/g°C.

Now substituting all values in (mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

(6.98)(91.29 - 33.54)(Cp) = (114.84)(33.54 - 24.37)(4.184)

solving, we get,

Cp = 10.93 J/g°C.

the specific heat of the metal is 10.93 J/g°C.

7 0

3 years ago