Answer:
The answer to the question is
The height of the mercury fluid column remain the same.
Explanation:
The pressure, P in a column of fluid of height, h is given by
P = (Density p)×(height of fluid column h)×(gravity g)
Therefore, when the diameter is doubled we have
Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence
P = p×h1×g = p×h2×g Therefore h1 = h2
The height of the fluid column remain the same
Answer:
Explanation:
We will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
Mᵣ: 18.02
2Na + H₂O ⟶ 2NaOH + H₂
m/g: 72.0
2. Moles of H₂O
3. Moles of Na
The molar ratio is 2 mol Na/1 mol H₂O.
Given:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
Assume that the gas ideally, thus, we can use the ideal gas equation:
PV = nRT
where R = 0.0821 L atm/mol K
123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T
solve for T
T = 295.72 K<span />
