So, could you use easily use chromic acid to visualize spots when you are monitoring a reaction to make a ketone out of an alcoh

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.

Answer: The percentage abundance for isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

  .....(1)

From the information provided:

Let the fractional abundance for isotope be 'x'

• For isotope:

Mass of isotope = 27.9769 amu

Percentage abundance of isotope = 92.22 %

Fractional abundance for isotope = 0.9222

• For isotope:

Mass of isotope = 28.9764 amu

Percentage abundance for isotope = 4.68%

Fractional abundance of isotope = 0.0468

• For isotope:

Mass of isotope = 29.9737 amu

Fractional abundance for isotope = x

• The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

To convert this fractional abundance into a percentage, multiply by 100:

This shows that the percentage abundance for isotope is 3.09 %.