a chemist uses hot hydrogen gas to convert chromium (iii) oxide to pure chromium. how many grams of hydrogen are needed to produ

Answer:

The solution to your inquiry is C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

Explanation:

Data

Q = 1.67 kcal

mass = 79.2 g

ΔT = 63.3°C

Formula

Q = mCΔT

Solving for C

C = Q/mΔT

Substituting values

C = 1.67/(79.2 x 63.3)

Simplifying

C = 1.67 / 5013.4

Final Result

C = 0.000333 kcal/g°C

or C = 0.333 cal/g°C

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Given data:

CO concentration in air = 15 ppm

Volume of air inhaled per breath = 0.50 L

Breaths taken per minute = 20

CO density = 1.2 g/L

Objective:

milligrams of CO inhaled over 6 hours

Clarification:

A concentration of 15 ppm of CO means that there are 15 liters of CO per 10⁶ liters of air

. Consequently, the volume of CO inhaled through 0.50 L of air is

= 15 L CO * 0.50 L air/10⁶ L air = 7.5 *10⁻⁶ L CO/breath

Next, considering there are 20 breaths in a minute,

the total number of breaths in 360 minutes (or 6 hours) will be

= 360 min * 20 breaths/1 min = 7200 breaths

Thus, the total volume of CO inhaled in that time frame is

= 7200 breaths * 7.50*10⁻⁶L/1 breath = 0.054 L

Given that the density of CO is 1.2 g/L

the mass of CO inhaled equals Density*Volume

= 0.054 * 1.2 = 0.0648 g = 64.8 mg

Thus, the mass of CO inhaled over 6 hours is 64.8 mg

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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