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madreJ
3 months ago
11

Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7

kj/mol, δh°f[co2(g)] = –393.5 kj/mol, δh°f[h2o(g)] = –241.8 kj/mol, how much energy is released (kj) when 8.30 g of butane is burned?
Chemistry
1 answer:
KiRa [2.9K]3 months ago
6 0

The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

The Δ H value for H2O (g) is -241.8 kJ/mol.

The mass of butane is 8.30 grams.

Butane has a molar mass of 58 g/mol.

Considering the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

To determine the Δ H° of the reaction:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

By substituting values, we find that

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, we will calculate how many moles of butane are in 8.30 grams.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Thus, the total heat released in the reaction is 380.14 kJ.

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it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine
alisha [2963]

Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ    

Where h is Planck's constant    = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ  

λ   = hc / E

λ   =  (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ   = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

6 0
3 months ago
Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the
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8 0
2 months ago
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A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
KiRa [2933]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product (CH3COOH)^{2} gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol (CH3COOH)^{2} / 2 mol CH3COOH)

= 0.005 mol (CH3COOH)^{2}

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

4 0
3 months ago
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