It can be anything with a big heat capacity like water
Answer:
V₂ = 107.84 L
Explanation:
Given data:
Initial volume = 100 L
Initial pressure = 80 KPa (80/101 =0.79 atm)
Initial temperature = 200 K
Final temperature =273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm
V₂ =21567 atm.L.K /200 K.atm
V₂ = 107.84 L
Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate

Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of

= 207+56+192 = 455 g/mol
Formula:

Solving:




Answer:
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B. 0.119 M</span>
Answer:
the results of an investigation thank me later
Explanation:
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C