Answer:
Refer to the explanation.
Explanation:
Formation reactions involve the creation of one mole of a compound from its elements in their standard states.
NaBr (s)
The equation for the standard formation is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
As per appendix C, the standard heat of formation for NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The equation for the standard formation is
S (s) + (3/2) O₂ (g) → SO₃ (g)
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ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The equation for the standard formation is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
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Answer:
THE MOLAR MASS OF XCL2 IS 400 g/mol
THE MOLAR MASS OF YCL2 IS 250 g/mol.
Explanation:
We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.
Number of moles = mass / molar mass
Molar mass = mass / number of moles.
For XCl2,
mass = 100 g
number of moles = 0.25 mol
Thus, molar mass = mass / number of moles
Molar mass = 100 g / 0.25 mol
Molar mass = 400 g/mol.
For YCl2,
mass = 125 g
number of moles = 0.50 mol
Molar mass = 125 g / 0.50 mol
Molar mass = 250 g/mol.
Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.
Vapor pressure refers to the force exerted by vapor or gas molecules above the surface of a liquid. It is inversely related to the concentration of solute particles; an increase in solute concentration results in a decrease in vapor pressure, and vice versa. For (a), it dissociates into two particles. In (b), the total count of particles from dissociation becomes 1 + 2, totaling three. For (c), dissociation yields 1 + 3 for a total of four particles. (d) Since sucrose is a covalent compound, it does not break apart into ions, so it remains as one particle. For (e), dissociation results in 1 + 1, equating to two particles.
Solution:
The gas's new temperature is 604K
Justification:
Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;
Step 1: Formulate the general gas law equation
P1V1/T1 = P2V2/T2
Step 2: Insert the values, converting as needed to standard units.
P1 = 0.800 atm
V1 = 0.180 L
T1 = 29°C = 273 + 29 = 302K
P2 = 3.20 atm
V2 = 90 mL = 90 * 10^-3 L = 0.09 L
Step 3: Solve for T2
The new gas temperature T2 is calculated as:
T2 = P2V2T1/(P1V1)
T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)
T2 = 86.976 / 0.144
T2 = 604K
The gas's new temperature is 604K.
