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Leno4ka [110]
3 years ago
5

8. Referring to the metabolic process involving glucose, calculate the maximum number of moles of ATP that can be synthesized fr

om ADP from the breakdown of one mole of glucose. (8 points) C6H12O6 + 6O2 → 6CO2 + 6H2O ΔGº = –2880 kJ/mol ADP + H3PO4 → ATP ΔGº = +31 kJ/mol
Chemistry
1 answer:
stepan [7]3 years ago
5 0

Answer:

93 moles

Explanation:

The breakdown of 1 mole of glucose is shown by the reaction shown below:

C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O\ \Delta G^0=-2880 kJ/mol

It means that 1 mole of breakdown of glucose produces 2880 kJ of energy.

The reaction of the phosphorylation of ADP to ATP is shown below as:

ADP+H_3PO_4\rightarrow ATP\ \Delta G^0=31 kJ/mol

It means that 1 mole of ADP needs 31 kJ of energy to convert into 1 mole of ATP

Which also means that:

31 kJ of energy is required to form 1 mole of ATP from ADP

1 kJ of energy is required to form \frac {1}{31} mole of ATP from ADP

2880 kJ of energy is required to form \frac {1}{31}\times 2880 mole of ATP from ADP

<u>Thus moles of ATP form ≅ 93 moles</u>

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A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
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Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

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