Methane (CH4, 16.05 g/mol) reacts with oxygen to form carbon dioxide (CO2, 44.01 g/mol) and water (H2O, 18.02 g/mol). Assume tha

Result:- 121 kPa.

Solution:- The parameters mass, temperature, and volume for carbon dioxide gas are given, and we are tasked with determining the pressure exerted by this gas on the container.

This is based on the ideal gas law equation, PV = nRT.

Where P stands for pressure in atm, V is the volume in liters, n is the gas moles, R is the universal gas constant with its value and T is the temperature in Kelvin.

Data provided:- mass = 88.0 g

T = 291 K

V = 40.0 L

P =?

We must convert mass into moles by dividing the mass by molar mass, where the molar mass of carbon dioxide is 44.01 grams per mole.

n = 2.00 mol

Reorganizing the equation for pressure gives us:

Let’s substitute the values and calculate for P.

P = 1.1946 atm

Since the question requests the answer in kPa, we need to convert atm to kPa.

1 atm = 101.325 kPa

Thus,

= 121 kPa

Consequently, the pressure this gas applies on the container is 121 kPa.

Answer:

The typical atomic weight of bromine is 79.9 amu.

Explanation:

Provided information:

Br⁷⁹ constitutes 55% of the sample

Br⁸¹ constitutes 45% of the sample

What is the average atomic weight of bromine?

Calculation formula:

Average atomic mass = [isotope mass × its proportion] + [isotope mass × its proportion] +...[ ] / 100

We can now insert the known values into the formula.

Average atomic mass = [55 × 79] + [81 × 45] / 100

Average atomic mass = 4345 + 3645 / 100

Average atomic mass = 7990 / 100

Average atomic mass = 79.9 amu

The typical atomic weight of bromine is 79.9 amu.

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

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.

Solution:

The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O

To find the moles of each element, we divide by their molar masses:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

Next, we find the mole ratio by dividing each by the smallest number of moles:

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Thus, the molecular formula is PbSO₄, representing lead sulfate.