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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

2 answers:

4 0

To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8%

Katen [24]3 years ago

4 0

If the actual yield of ferrous sulfide is 220.0 g and its theoretical yield is 275.6 g, the percent yield of ferrous sulfide is 79.8%. Make a proportion. 220.0 g is x percent. 275.6 g is 100%. 220.0 g : x = 275.6 g : 100%. After crossing the products: x = 220.0 g * 100% : 275.6 g = 79.8%

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How many atoms are in each compound ?

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Compounds are made of two or more atoms of different elements, such as water (H2O) and methane (CH4). Atoms are not drawn to scale. Molecules of compounds have atoms of two or more different elements.

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2 years ago

A student balances the following redox reaction using half-reactions.

ICE Princess25 [194]

Answer: 6.

Explanation:

1) Aluminum

$Al^0-3e^----\ \textgreater \ Al^{3+}$

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

$Mn^{2+}+2e^{-}---\ \textgreater \ Mn$

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

$2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}$

4) Net equation

Add the two half-equations:

$2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}$

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.

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2 years ago

Read 2 more answers

What is the importance of a special habitat?

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They meet all the environmental conditions an organism needs to survive

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1 year ago

Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First we convert 54.0 g of propane (C₃H₈) into moles, using its molar mass:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we convert 1.23 moles of C₃H₈ into moles of CO₂, using the stoichiometric coefficients:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We convert 3.69 moles of CO₂ into grams, using its molar mass:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And apply the given yield:

162.36 g * 64.0/100 = 103.9 g

7 0

2 years ago

What is the IUPAC name for this product?

mihalych1998 [28]

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

In the given product total four carbon atoms are present and between all of them single bonds are present.
In the second carbon atom, chlorine group is present.
During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

To know more about IUPAC nomenclature, visit the below link:

brainly.com/question/26635784

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1 year ago