Answer:
Freezing T° of solution = - 4.52°C
Explanation:
ΔT = Kf . m . i
That's the formula for colligative property about freezing point depression.
Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.
Li₂O → 2Li⁺ + O⁻²
3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.
m is molality → 0.811 m, this is data
Kf →Cryoscopic constant, for water is 1.86 °C/m
and ΔT = Freezing T° of pure solvent - Freezing T° of solution
We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3
Freezing T° of solution = - 4.52°C
Answer:
Your answer choice is D
Explanation:
I just got it right on USATestprep :))
Heat flows from our body to iron
The physical changes are:
- condensation of water vapors
- carving wax
- slicing a banana tearing
- aluminum foil
The chemical changes are:
- bleaching hair
- silver tarnishing,
- frying an egg, and
- burning paper
<h3>What are physical and chemical changes?</h3>
Physical changes are changes which occur in a substance which may alter the physical properties of the substance but does not affect the chemical properties of the substance. Physical changes are easily reversible
In physical changes, no new substances are formed. Example of a physical change is the melting of ice or candle wax.
Chemical changes are changes which alter the chemical properties of a substance and in which new substances are formed. Example of chemical changes are; the rusting of iron, bleaching hair silver tarnishing, frying an egg, and burning paper.
Chemical changes are not easily reversible.
In conclusion, physical changes do not produce new substances while chemical changes produce new substances.
Learn more about physical and chemical changes at: brainly.com/question/18526179
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Answer:
V₂= 1.9 L
Explanation:
Given data:
Initial volume of gas = 2.20 L
Temperature = 30°C (30+273 = 303 K)
Initial pressure = 735.43 mmHg (735.43 /760 = 0.97 atm)
Final volume of gas = ?
Final temperature = standard = 273 K
Final pressure = standard = 1 atm
Solution:
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 = P₁V₁ T₂/ T₁ P₂
V₂ = 0.97 atm × 2.20 L × 273 K / 303 K × 1 atm
V₂= 582.58 atm .L. K / 303 k.atm
V₂= 1.9 L
