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vichka [17]
3 years ago
7

for a theoretical yield of 22 g and actual yield of 13 g, calculate the percent yield for a chemical reaction

Chemistry
1 answer:
sammy [17]3 years ago
8 0

Answer:

59.09%

Explanation:

13/22*100=59.09%

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Answer:

(CH₃)₃COCH3₃ and (CH₃)₂CHOCH₂CH₃

Explanation:

Isomers are compounds which have the same molecular formula. Constitutional isomers have different connectivity; the atoms are connected in different ways.

1. (CH₃)₃COCH₃

2. (CH₃)₂CHOCH3₃

3. (CH₃)₂CHOCH₂CH₃

Molecules 1 and 3 have the same formula (C₅H₁₂O) and are isomers. Molecule 2 is not an isomer. From the structural formula, it is clear that Molecules 1 and 3 have different connectivity.

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Calculate the kilograms of iron that would be produced from 1340 g of calcium carbonate.
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1.340kg

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In a sample of solid ba(no3)2 the ratio of barium ions to nitrate ions is
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<span>In a sample of solid Ba(NO3)2 the ratio of barium ions to nitrate ions is would be one is to 2 or 1:2. Barium ion has a formal charge of positive two which means that it needs two ions which has a formal charge of negative one or 1 ion with the formal charge of negative two. However, for this case, it is bonded to a nitrate ion which has a formal charge of negative one. Therefore, it needs two nitrate ions so that for every 1 atom of barium ion, we need two ions of nitrate ions.</span>
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Are the particles moving faster or slower as time goes on? (Diagram F)
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3 years ago
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 %

3 0
3 years ago
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