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Nonamiya [84]
3 years ago
13

After Callie obtained her original results, she wanted to dig deeper. She determined that the germinating corn seed had utilized

100 molecules of glucose in the first few seconds that she performed the experiment. If these data are accurate, how many carbon dioxide molecules would Callie expect to have been released as a waste during this same amount of time? Explain how you arrived at your answer. The formula for cellular respiration is shown below.
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
Chemistry
2 answers:
Airida [17]3 years ago
8 0

Answer:

Callie expect 600 molecules of CO2 to have been released as a waste during the same amount of time.

Explanation:

During cellular respiration 1 molecule of glucose undergoes oxidation to form 6 molecules of CO2 as a waste product.

     According to the question callie determined that the germinating corn seed had utilized 100 molecules of glucose.

    So 100 molecules of glucose will release 100×6=600 molecules of CO2 as a waste product.

blondinia [14]3 years ago
4 0

Answer: 600 molecules of CO2

Explanation:

C6H12O6 + 6O2 → 6CO2 + 6H2O + energy

From the above balanced equation, 1 mole of glucose produces 6 moles of CO2.

Mathematically,

1 mole of C6H12O6 = 6 moles of CO2

Let,

100 moles of C6H12O6 = x mole of CO2

By simply cross multiply,

x mole of CO2 = (6 ×100)/1 mole

x mole of CO2 = 600 mole

Callie should expect 600 molecules of carbon dioxide to have been released as a waste during this same amount of time.

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3 years ago
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
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Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

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\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

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d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

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3 years ago
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Answer:

solution is clear solution while colloidal is between the solution and suspension. And in suspension particles are suspended.

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In solution light can be passed without any scattering of light from solute particles while suspension is cloudy and having larger particle size than colloids, if suspension stands for a while particles will settle down easily.

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6 0
3 years ago
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Hey!!! Chemistry Question about MOLES... Please help...
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Answer:

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Explanation:

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2 years ago
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