The addictive quality is a reduction in vapour pressure.
The vapour pressure at a liquid's normal boiling point is the same as the ordinary atmospheric pressure, which is 1 atmosphere, 760 Torr, 101.325 kPa, or 14.69595 psi.
The pressure that results from liquids evaporating is known as vapour pressure. Surface area, intermolecular forces, and temperature are three often occurring variables that affect vapour press.
lower vapour pressure
raising the boiling point
Low-temperature depression
Osmotic force
They are all dependent on the solute; when you increase the solute, the colligative property and the ratio you added may change.
The Van't Hoff Factor is another option to examine (i). the number of dissolved ions. The colligative property will be further altered if the solute is ionic.
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Answer:
V ∝ n
Step-by-step explanation:
Suppose that pressure and temperature are constant.
If you try to force more molecules of air into a balloon, the balloon will expand.
This is an example of <em>Avogadro's Law</em>: the volume of a gas is directly proportional to the number of moles (particles).
V ∝ n
Answer:62.66°C or 235.66K
Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.
17*4.184=71.128kJ.
71.128kJ=mcpT
71.128kJ=245*4.187*(T-Tm)
Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.
71128J=245*4.187*(305-Tm)
71128=312873.575-1025.815Tm
1025.815Tm=312873.575-71128
1025.815Tm=241745.58
Tm=241745.58/1025.815
Tm=235.66K
Answer:
The correct answer is c) 134L
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.
1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K
V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm
V = 133, 86828 l
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Explanation:
