A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L. (a) What is the pH o

Answer: The mass of Si in kilograms is,

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si =

Density of Fe =

Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms =

Si concentration in kilograms =

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms =

Si concentration in kilograms =

Hence, the mass of Si in kilograms is,

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Identifying the given parameters from the question;

Vapor pressure = 94.4 mm of Hg

The reaction for vaporization is expressed as;

C₆H₆(l) ⇄ C₆H₆(g)

The equilibrium in terms of activities can be defined as:

K = a(C₆H₆(g)) / a(C₆H₆(l))

The activity for pure substances equals one:

a(C₆H₆(l)) = 1

For an ideal gas phase, activity is approximated as the ratio of partial pressure to total pressure. Under standard conditions:

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg is the standard pressure

Thus, we find;

K = 94mmHg / 760mmHg = 0.12421

The formula for Gibbs free energy is:

ΔG = - R·T·ln(K)

Here, R represents the gas constant = 8.314472J/molK

Consequently, the ΔG° for the vaporization of benzene is calculated as:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

The change in Gibbs free energy for the reaction is determined by the difference between the Gibbs free energy of formation of the products and reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = ΔvG° + ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol