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2 years ago

The heat of combustion of oleic

Chemistry

1 answer:

Karolina [17]2 years ago

3 0

Answer: The heat of formation of oleic acid is -94.12 kJ/mol

Explanation:

We are given:

Heat of combustion of oleic acid = $-1.11\times 10^4kJ/mol$

The chemical equation for the combustion of oleic acid follows:

$C_{18}H_{34}O_2(l)+\frac{51}{2}O_2(g)\rightarrow 18CO_2(g)+17H_2O(g);\Delta H^o=-1.11\times 10^4kJ$

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(18\times \Delta H^o_f_{(CO_2(g))})+(17\times \Delta H^o_f_{(H_2O)})]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=-1.11\times 10^4kJ$

Putting values in above equation, we get:

$-1.11\times 10^4=[(18\times (-393.51))+(17\times (-241.82))]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times 0)]\\\\\Delta H^o_f_{(C_{18}H_{34}O_2(l))}=-94.12kJ/mol$

Hence, the heat of formation of oleic acid is -94.12 kJ/mol

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2 years ago

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3 years ago

When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g

Serga [27]

Answer: The mass of iron produced will be 77.6 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For FeO:

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol$

For aluminium:

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol$

The given chemical reaction follows:

$3FeO+2Al\rightarrow 3Fe+Al_2O_3$

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = $\frac{3}{2}\times 0.93=1.395mol$ of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = $\frac{3}{2}\times 0.93=1.395moles$ of iron metal

Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

$1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g$

Hence, the mass of iron produced will be 77.6 grams

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3 years ago

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera

Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

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2 years ago

Change the bond between the two carbon atoms in each molecule to a double or triple bond as needed to complete the structure. If

Georgia [21]

Answer:

Your question is complex, because I think you wrote it wrong.

Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.

This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.

Simple junctions talks about compound organisms called ALKANS.

The double unions, in organic these compounds are called as ALQUENOS.

And as for the tertiary unions, the organic chemistry names them as ALQUINOS.

These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.

Explanation:

In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties

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2 years ago