Answer:
The factor that will change the volume of the diver's lungs upon reaching the surface is 4
Explanation:
Given data:
Pressure increases 1 atm = 101.325 kPa
34 ft = 10.3632 m
Depth of 102 ft = 31.0896 m
Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?
The pressure at 31.0896 m:
The factor will the volume of the diver's lungs change upon arrival at the surface:
Q = mcΔT
m = 75 g
c = 4.25 J/gram°C
q = 2450 J
q/mc = ΔT
2450/(75 • 4.25) = 7.7°C
The temperature increased by 7.7°C.
Answer: 167 g
Explanation:
1) The depression of the freezing point of a solution is a colligative property ruled by this equation:
ΔTf = i × m × Kf
Where:
ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.
i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1
Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C
2) Calculate the molality (m) of the solution
ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m
3) Calculate the number of moles from the molality definition
m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent
moles of solute = 2.69 m × 1.00 kg = 2.69 moles
4) Convert moles to grams using the molar mass
molar mass of C₂H₆O₂ = 62.07 g/mol
mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g
Explanation:
here's the answer to your question
Answer:
Sulfur's Atomic #: 16
Fluorine's Atomic #: 9
Antimony's Atomic #: 51
Silver's Atomic #: 47
Rubidium's Atomic #: 37
Copper's Atomic #: 29
Tin's Atomic #: 50
Mercury's Atomic #: 80
Promethium's Atomic #: 61
