The graph shows a sample of gas when it is hot, cold, and at room temperature.

2C6H14 + 13O2 ---> 6CO2 +14H2O

Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

Thus, 86.17 g of C6H14 corresponds to 1 mole.

                                  2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation        2 mol                            6 mol
according to the question    1 mol                            3 mol

To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2

1. The luminol stock solution has a molarity of 1.431 M.

2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.

3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.

Additional Information

Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.

Key concepts in stoichiometry include:

• 1. Relative atomic mass
• 2. Relative molecular mass

This refers to the relative atomic mass of a molecule.

• 3. Mole

A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.

1 mole = 6.02 × 10²³ particles.

The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.

Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.

To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.

Thus, the molarity is calculated as:

• 1. Moles of Luminol

- the relative molecular mass of Luminol:

= 8.C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams/mol.

Thus, we have:

moles = grams / relative molecular mass.

moles = 0.1073.

2. Molarity (M)

M = moles / volume

M = 1.431.

• b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:

moles = M × volume

moles = 6 × 10⁻² × 2

moles = 0.12.

• c. The molarity of the stock solution (Part A) is 1.431 M.

The diluted solution (Part B) contains 0.12 moles of luminol.

To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):

volume = moles / M

volume = 0.08386 L = 83.86 mL.

Further Learning

moles of water you can generate

the amount of each atom in the chemical's formula

the proportion of hydrogen to oxygen atoms in 2 L of water

Keywords: mole, volume, molarity, Luminol, relative molecular mass

Different wavelengths are involved.

Explanation:

When magnesium ignites with a bright white flame, it indicates that various wavelengths are related to the electron transitions occurring in the magnesium atom.

• Upon combustion, the electrons within the atom become excited.
• They emit characteristic light that corresponds to their energy levels.
• White light consists of a mix of different wavelengths.
• Seeing white light implies that multiple wavelengths combined are responsible for the observed emission.

Learn more:

Spectrum

Answer:

The original halide's formula is SrCl₂.

Explanation:

• The chemistry reaction's balanced equation is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.

• Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.

• The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.

• The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.

• n = mass / molar mass, thus n =  4.11 x 10⁻³ moles and mass = 0.652 g.

• The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.

• The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).

• Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.

• This identifies the atomic mass of Cl.

• Consequently, the original halide's formula is SrCl₂.

1 percent of the solar radiation