Methane gas is expanded from an initial volume of 20.5 L at 0.92 atm at 23oC to a final volume of 34.6 L. During the expansion t

<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>

Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ    

Where h is Planck's constant    = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ  

λ   = hc / E

λ   =  (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ   = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

Answer: The correct selection is (b).

Explanation:

The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.

This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.

Therefore, a significant amount of energy is needed to dislodge the valence electrons.

The electronic configuration for helium is . Hence, due to its fully occupied valence shell, it exhibits greater stability.

Consequently, a large amount of energy is needed to remove an electron from a helium atom.

In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.

In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.

1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).