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Gekata [30.6K]
3 years ago
5

Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer:

We need 67.2 grams of 02. There  will be produced 92.421 grams of CO2

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6 (aq) + 6O2 (g) = 6CO2 (g) + 6H2O (l)

<u>Step 2</u>: Data given

We can see that for 1 mole C6H12O6 consumed, we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O

The mass of glucose is 63.0g

The molar mass of glucose is 180.156 g/mole

The molar mass of O2 is 32 g/mole

The molar mass of CO2 is 44.01 g/mole

<u>Step 3: </u>Calculate number of moles of glucose

Number of moles of glucose = mass of glucose / molar mass of glucose

Number of moles of glucose = 63 grams /180.156 g/moles = 0.35 moles

<u>Step 4:</u> Calculate number of moles of oxygen ( and CO2)

For 1 mole of glucose we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O

So for 0.35 moles of glucose we need 6*0.35 = 2.1 moles of O2 to produce 2.1 moles of CO2 and 2.1 moles of H2O

<u>Step 5:</u> Calculate mass of oxygen ( and CO2)

mass of oxygen = number of moles of oxygen * Molar mass of oxygen

mass of oxygen = 2.1 moles * 32 g/mole = 67.2 g O2 (≈ 67 grams)

mass of CO2 = number of moles of CO2 * Molar mass of CO2

mass of CO2 = 2.1 moles * 44.01 g/mole = 92.421 grams

We need 67.2 grams of 02. There  will be produced 92.421 grams of CO2

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Quartenary.

Explanation:

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Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p
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% abundance of isotope 2 = 0.063 % = \frac{0.063}{100}=6.3\times 10^{-4}

Mass of isotope Ar- 40 = 39.96 amu

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Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

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Explanation:

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Answer:

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example  when  using  copper  it is  written  as  follows

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cu (s)  -->  CU2+   +2e    -0.34v

Explanation:

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