Answer:
0.045 moles
Explanation:
Given data:
Moles of CO = ?
Volume of gas = 1 L
Temperature and pressure = standard
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1 atm× 1L / 0.0821 atm.L/mol.K × 273 K
n = 1 atm.L / 22.41 atm.L/mol
n = 0.045 mol
I think the answer is OPC
The answer is would be B! :)
Answer:
The answer to your question is 510.7 kJ
Explanation:
Data
Q = ?
Volume = 2 L
Temperature 1 = T1 = 19°C
Temperature 2 = T2 = 80°C
Specific heat = 4.186 J/kg°C
Density of water = 1 g/ml
Equation
Q = mC(T2 - T1)
Process
1.- Calculate the mass of water
mass = density x volume
mass = 1 x 2000
mass = 2000 g
2.- Substitute values in the Heat formula
Q = (2000)(4.186)(80 - 19)
-Simplification
Q = (2000)(4.186)(61)
-Result
Q = 510692 J or 510.7 kJ
