Response:
K = 6.5 × 10⁻⁶
Detailed Explanation:
C₅H₆O₃ ⇄ C₂H₆ + 3CO
Apply PV=nRT to determine the initial pressure of C₅H₆O₃
P (2.50) = (0.0493) (0.08206) (473)
P = 0.78atm
C₅H₆O₃ ⇄ C₂H₆ + 3CO
0.78atm 0 0
0.78 - x x 3x
1.63atm = 0.78 - x + x + 3x
P(total) = 0.288atm
C₅H₆O₃ = 0.78 - 0.288
= 0.489atm
C₂H₆ = 0.288atm
CO = 0.846atm
= 0.379
= 6.5 × 10⁻⁶
Answer:
around 40
Explanation:
The included diagram shows 5 hydrophobic molecules, each surrounded by 9 water molecules. Therefore, there are "around 40" water molecules that are in contact with the hydrophobic molecules.
The enthalpy of hydration for copper sulfate is -1486.62 kJ/mol, indicating that 1486.62 kJ of energy is absorbed by a mole of copper sulfate during its hydration. Step 1: Calculate the energy released per mole of dissolved substance (Eq. 1). If 0.102 moles release 55.51 kJ, then 1 mole corresponds to 541.85 kJ/mol. Therefore, ΔH = -541.85 kJ/mol. Step 2: Identify the energy absorbed by dissolved substance (Eq. 2). When 0.101 moles absorb 95.31 kJ, 1 mole will absorb 944.77 kJ/mol, thus ΔH = 944.77 kJ/mol. Step 3: Subtract Eq. 2 from Eq. 1. Thus, ΔH = -541.85 kJ/mol (Eq. 1) and ΔH = 944.77 kJ/mol (Eq. 2), leading to ΔH = -541.85 - 944.77, so ΔH = -1486.62 kJ/mol.
Density is calculated as Mass divided by Volume.
Volume can also be expressed as Mass divided by Density.
Volume becomes 2 divided by 128.
This results in Volume equal to 1/64.
Therefore, each side measures 4 cm.
