CAN SOMEONE TELL ME THE answer for this ?
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
Explanation:
It is given that,
Mass of the brick, m = 1.15 kg
Radius of the circle, r = 1.44 m
The cable will break if the tension exceeds 43.0 N
Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,
v = 6.30 m/s
So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.