Which device is used to measure interior diameter of a water pipe and in which units are the callibrations
2 answers:
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;
The resistance of the loop = 0.7 Ω
The induced current is calculated as;