The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o

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.

For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).

From the provided data, the unknown mixture was composed of salt, salicylic acid, and sand. It is understandable that the student suspected the presence of sand, yet scientific experimentation must verify such assumptions. The test involving salt and salicylic acid reveals that salt dissolves in water, while salicylic acid is only slightly soluble, and sand does not dissolve at all. By introducing the unknown into water, the salt would dissolve first, followed by the partial dissolution of salicylic acid. Heating the mixture could allow for the evaporation of salicylic acid, resulting in the remaining salt. If traces of sand were observed in the dissolved sample, it could suggest contamination.

Answer:

The categorization of strong, weak, and non-electrolytes is detailed below, based on the examples presented in the question.

Explanation:

A strong electrolyte fully dissociates or nearly so in an aqueous environment; typically, strong acids, bases, and salts fall under this category. Examples of strong electrolytes include:

• Hydrochloric acid, HCl
• Calcium hydroxide, Ca(OH)2
• Potassium chloride, KCl

A weak electrolytepartially ionizes in solution; weak acids and bases are primary instances. Examples consist of:

• Methylamine, CH3NH2
• Hydrofluoric acid, HF

A non-electrolytedoes not dissociate in an aqueous medium. Examples of non-electrolytes are:

• Sucrose, C12H22O11
• Methanol, CH3OH