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Gala2k [10]
3 years ago
11

Calculate the approximate enthalpy change, δhrxn, for the combustion of one mole of methane a shown in the balanced chemical equ

ation: ch4+2o2→2h2o+co2 use the values you calculated in parts a, b, c, and d, keeping in mind the stoichiometric coefficients
Chemistry
1 answer:
Elena L [17]3 years ago
4 0
To solve for the enthalpy of reaction, we apply the Hess's Law.

ΔHrxn = ∑(ν×Hf of products) - ∑(ν×Hf of reactants)
where
v is the stoichiometric coefficient determined from the balanced reaction
Hf is the standard heat of formation; these are empirical values:
*For CH₄: Hf = <span>−74.87 kJ/mol
*For O</span>₂: Hf = 0
*For CO₂: <span>-393.5 kJ/mol 
*For H</span>₂O: <span>-241.82 kJ/mol

</span>ΔHrxn = [(2*-241.82 kJ/mol)+(1*-393.5 kJ/mol)] - [(1*−74.87 kJ/mol)+(2*0 kJ/mol)] =<em> -802.27 kJ/mol</em>
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Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }

6,61 = √M₂

44g/mol = M₂

<em>The molar mass of the gas is 44 g/mol</em>

<em></em>

I hope it helps!

4 0
3 years ago
6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) soluti
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Answer:

1.047 M

Explanation:

The given reaction:

6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}

For dichromate :

Molarity = 0.254 M

Volume = 15.8 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

Thus, moles of dichromate :

Moles=0.254 \times {15.8\times 10^{-3}}\ moles

Moles of dichromate = 0.0040132 moles

1 mole of dichromate react with 6 moles of iron(II) solution

Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

Moles of iron(II) solution = 0.02408 moles

Volume = 23 mL = 0.023 L

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>

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Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. Therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

Explanation:

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