Answer:
A Broad View of Concept Modeling
Establishing Key Relationships
Levels of Concept Modeling
Explanation:
Answer:
B = - 0.0326 dm³/mol
Explanation:
virial eq until second term:
∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm
∴ T = 200°C = 473 K
∴ Vm = 3.90 dm³/mol
∴ R = 0.08206 dm³.atm/K.mol
⇒ PVm / RT = 1 + B/Vm
⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm
⇒ 0.99164 = 1 + B/Vm
⇒ B/Vm = - 8.357 E-3
⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )
⇒ B = - 0.0326 dm³/mol
Answer:
V = 1.434 L
Explanation:
Given data:
Mass of argon = 4.24 g
Temperature = 58.2 °C
Pressure = 1528 torr
Volume = ?
Solution:
58.2 °C = 58.2 + 273 = 331.2 K
1528/760= 2.01 atm
<em>Number of moles:</em>
Number of moles = mass/molar mass
Number of moles = 4.24 g / 39.948 g/mol
Number of moles = 0.106 mol
<em>Volume:</em>
PV = nRT
V = nRT/P
V = 0.106 mol ×0.0821. atm. L. mol⁻¹. K⁻¹ × 331.2K/ 2.01 atm
V = 2.88 atm L/ 2.01 atm
V = 1.434 L
Answer:
For terrestrial animals, grazing is normally distinguished from browsing in that grazing is eating grass or forbs, and browsing is eating woody twigs and leaves from trees and shrubs. Grazing differs from true predation because the organism being grazed upon is not generally killed.
Explanation:
I believe the criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the <span>particles remain suspended for an extended period of time. I hope it helps you.</span>
