Chemistry
16.02.2023 14:30
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List examples of organisms from the movie that represent each of the six kingdoms.

List examples of organisms from the movie that represent each of the six kingdoms. Monera (Archaebacteria/Eubacteria): Protista: Fungi: Plantae: Animalia: (finding nemo)
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cganderson04
cganderson04
4,5(41 marks)

Animalia: Nemo

Plantae: coral

Fungi: algae

Explanation:

I only know 3 sorry

cman8228
cman8228
4,9(12 marks)

Explanation:

For 1:

We are given:

Trial 1:

Mass of empty crucible with lid = 26.698 g

Mass of Mg O + crucible + lid = 27.291 g

Actual yield of MgO = (27.291 - 26.698) g = 0.593 g

Trial 2:

Mass of empty crucible with lid = 26.687 g

Mass of Mg O + crucible + lid = 27.273 g

Actual yield of MgO = (27.273 - 26.687) g = 0.586 g

For 2:

To calculate the theoretical yield of magnesium oxide, we first need to find the actual yield of magnesium for both the trials.

Trial 1:

Mass of empty crucible with lid = 26.698 g

Mass of Mg + crucible + lid = 27.060 g

Actual yield of Mg = (27.060 - 26.698) g = 0.362 g

Trial 2:

Mass of empty crucible with lid = 26.698 g

Mass of Mg + crucible + lid = 27.046 g

Actual yield of Mg = (27.046 - 26.698) g = 0.359 g

Now, we calculate the number of moles of magnesium for both the trials.

Molar mass of magnesium = 24.31 g/mol

Trial 1:

\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=\frac{0.362g}{24.31g/mole}=0.015moles

Trial 2:

\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=\frac{0.359g}{24.31g/mole}=0.0148moles

Then, we calculate the number of moles of MgO for both the trial.

The balanced chemical equation for the formation of magnesium oxide follows:

2Mg+O_2\rightarrow 2MgO

Trial 1:

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.015 moles of Mg will produce = \frac{2}{2}\times 0.015=0.015 moles of MgO.

Trial 2:

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.0148 moles of Mg will produce = \frac{2}{2}\times 0.0148=0.0148 moles of MgO.

Now, calculating the theoretical yield of magnesium oxide.

Trail 1:

\text{Theoretical yield of MgO}=\text{Moles of MgO}\times \text{Molar mass of MgO}

\text{Theoretical yield of MgO}=(0.015mole)\times (40.3g/mole)=0.6045g

Trial 2:

\text{Theoretical yield of MgO}=\text{Moles of MgO}\times \text{Molar mass of MgO}

\text{Theoretical yield of MgO}=(0.0148mole)\times (40g/mole)=0.5964g

For 3:

To calculate the percentage yield, we use the formula:

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

Trial 1:

Actual yield of MgO = 0.593g

Theoretical yield of MgO = 0.6045 g

Putting values in above equation, we get:

\%\text{ yield of MgO}=\frac{0.593g}{0.6045g}\times 100=98.09\%

Trial 2:

Actual yield of MgO = 0.586 g

Theoretical yield of MgO = 0.5964 g

Putting values in above equation, we get:

\%\text{ yield of MgO}=\frac{0.586g}{0.5964g}\times 100=98.25\%

For 4:

To calculate the average yield, we add on total number of yields and divide it by the number of yields.

\text{Average percent yield of MgO}=\frac{\%\text{ yield of MgO for trial 1}+\%\text{ yield of MgO for trial 2}}{2}

\text{Average percent yield of MgO}=\frac{98.09+98.3}{2}=98.195\%

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