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2 years ago

In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?

1 answer:

5 0

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

$heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }$

heat= 4,707.74 kJ

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

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Why is gold usually found in its pure form

attashe74 [19]

gold is usually found in pure form because it is not reacting with other chemicals naturally.

5 0

3 years ago

Balance the following equation;C<img src="https://tex.z-dn.net/?f=C5H12%28g%29%2BO2%28g%29%3DC02%28g%29%2BH2O%28l%29" id="TexFor

Bumek [7]

put 8 in front of the oxygen in the reactants side to make it 16 molecules then put a 5 in front of the co2 in the product side to balance the carbon atoms then put a 6 in front of the H20 on the product side this balances both the hydrogen and oxygen atoms here is a representation

C5H12(g)+8O2(g)=5CO2(g)+6H20

4 0

2 years ago

slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

Answer:

For A: The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

Explanation:

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

For A: Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

6 0

3 years ago

Which metal atoms can form ionic bonds by losing electrons from both the outermost and next to outermost principal energy levels

RoseWind [281]

Full question options;

(Fe, Pb, Mg, or Ca)

Answer:

Iron - Fe

Explanation:

We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.

Pb, Mg, and Ca only lose their outermost electrons to form the following ions;

Pb2+, Mg2+, and Ca2+.

This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.

Iron on the other hand has the following electronic configurations;

Fe: [Ar]4s2 3d6

Fe2+: [Ar]4s0 3d6

Fe3+: [Ar]4s0 3d5

This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.

5 0

3 years ago

Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun

Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂= $\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}$ =0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

$\frac{0.92678}{0.0309}$ =Si³⁰

Si³⁰=29.99288

8 0

3 years ago