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kozerog [31]
2 years ago
13

Stearic acid (C18H36O2) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. I

t is used to make cosmetics, ointments, soaps, and candles and is found in animal tissue as part of many saturated fats. In fact, when you eat meat, you are ingesting some fats containing stearic acid. Determine the LaTeX: \DeltaΔHrxn for this combustion given the following information: LaTeX: \DeltaΔHf of stearic acid = -948 kJ/mol, LaTeX: \DeltaΔHf of CO2 = -394 kJ/mol, LaTeX: \DeltaΔHf of water = -242 kJ/mol. Calculate the heat (q) released in kJ when 262 g of stearic acid reacts with 914.5 g of oxygen. Enter a positive number to 0 decimal place.
Chemistry
1 answer:
Paha777 [63]2 years ago
6 0

Answer:

There is 9671  Kj of heat released

Explanation:

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

C18H36O2(s) + 26O2(g) --> 18CO2(g)+18H2O(g)

This means for 1 mole of C18H36O2 consumed there is 26 moles of O2 needed to produce 18 moles of CO2 and 18 moles of H2O.

<u>Step 2:</u> Calculate the heat of combustion

ΔH (combustion) = [18*(ΔHf of CO2) + 18*(ΔHf of H2O)] - [1*(ΔHf of C18H36O2) + 26*(ΔHf of O2)]

ΔH (combustion) = [18*(-394 kJ/mol) + 18*(-242 kJ/mol)] - [1*(-948 kJ/mol) + 26*(0 kJ/mol)]

ΔH (combustion) = [(-7092 kJ/mol) + (-4356 kJ/mol)] - (-948 kJ/mol)

= -10500 kJ/mol

ΔH (combustion) = heat released / number of moles of stearic acid

<u>Step 3:</u> Calculate moles of stearic acid

moles of stearic acid = mass / Molar mass of stearic acid

moles of stearic acid = 262g / 284.48 g/mole = 0.921 moles

<u>Step 4:</u> Calculate moles of oxygen

moles of O2 = 914.5 / 32g/mole

moles of O2 = 28.578125 moles

Stearic acid is the limiting reactant: it will <u>completely react</u>

There will react 26*0.921 mole = 23.946 mole of O2

This means there will remain 4.63 moles of O2

<u>Step 5:</u> Calculate heat released

q = (ΔH combustion) * (moles of stearic acid) = (-10500 kJ/mol) * (0.921 moles) = 9671 Kj

There is 9671  Kj of heat released

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Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.

Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.

Thus;

Ca(CH3COO)2(aq)  ----> Ca^2+(aq) + 2CH3COO^-(aq)

It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.

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Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith
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The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

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