Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
Answer:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Explanation:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Answer:
D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted
Explanation:
access-list 90
deny 10.168.7.0 0.0.0.255
permit 10.168.7.10
permit 10.168.7.11
permit 10.168.7.12
deny any
