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AfilCa [17]
4 years ago
6

Use Hess's law and the following equations to calculate the ΔHreaction for the reaction CO(g) + 3H2(g) CH4(g) + H2O(g). (Show yo

ur work.) (4 points) • C(s) + O2(g) CO(g) ΔH = –110.5 kJ • C(s) + 2H2(g) CH4(g) ΔH = –74.85 kJ • H2(g) + O2(g) H2O(g) ΔH = –241.83 kJ
Chemistry
1 answer:
Vesna [10]4 years ago
6 0

Answer: The enthalpy of reaction is, -206.18 kJ

Explanation:-

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The final reaction is

CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=-110.5kJ

(2) C(s)+2H_2(g)\rightarrow CH_4(g)    \Delta H_2=-74.85kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)    \Delta H_3=-241.83kJ

Now adding (2) and (3) and subtracting (1) , we get :

\Delta H_{rxn}=(\Delta H_2+\Delta H_3)-\Delta H_1

\Delta H_{rxn}=((-74.85)+(-241.83))-(-110.5)

\Delta H_{rxn}=-206.18kJ

Therefore, the enthalpy of reaction is, -206.18 kJ

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The earths rotation. The earth rotates and as it rotates, the sun moves across the sky

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3 years ago
What is the orientation of the two unhybridized p orbitals on Be with respect to the two Be−F bonds?
Len [333]

Answer:

b) Both p orbitals are perpendicular to the F−Be−F bond axes.

Explanation:

Be has 2 electrons in its valence shell, subshell s is fulfilled, so it has no unpaired electrons in its ground state to make bonds with F. So, it can promote the electrons to the 2p orbital and will having sp hybridization.

The bond between the orbitals sp and the p orbital of F are in opposite directions but the same ax. The two bonds are equivalent, and the molecule had a linear geometry. The two unhybridized p orbitals on Be are vacant, and so they are perpendicular to the F-Be-F bond axes.

7 0
4 years ago
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. if 4.40 mol of m
Georgia [21]
The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
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8 0
4 years ago
At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution ha
brilliants [131]

Answer:

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

<em>1.66x10⁻⁴ = [OH⁻]</em>

And [M²⁺] is the half of [OH⁻], <em>[M²⁺] = 8.30x10⁻⁵</em>

<em />

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

7 0
3 years ago
Write a resonance form for pyrrole in which nitrogen has a formal charge of 1. Are comparable resonance forms possible for pyrid
Svetach [21]

Answer:

See explanation and image attached

Explanation:

Here attached are resonance forms of pyrrole and pyridine. The images were obtained from quora and researchgate respectively.

Now, we can see that in the resonance forms of pyrrole, the nitrogen atom in the heterocycle has a formal charge of +1. However, in the six membered pyridine hetrocycle, the nitrogen atom may have a formal charge of -1 or +1 as shown in the canonical structures attached. The structures in which nitrogen has a +1 formal charge in pyridine are comparable to structures obtained from pyrrole. These structures have less contribution to the structure of pyridine.

3 0
3 years ago
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