0.01 m
< 0.03 m
< 0.04 m urea
As molal concentration rises, so does freezing point depression. It can be expressed mathematically as ΔTf = Kfm.
<h3>What is Colligative Properties ?</h3>
- The concentration of solute particles in a solution, not the composition of the solute, determines a colligative properties .
- Osmotic pressure, boiling point elevation, freezing point depression, and vapor pressure reduction are examples of ligand-like properties.
<h3>What is freezing point depression?</h3>
- When less of another non-volatile material is added, the temperature at which a substance freezes decreases, a process known as Freezing-point depression.
- Examples include combining two solids together, such as contaminants in a finely powdered medicine, salt in water, alcohol in water.
- An significant factor in workplace safety is freezing points.
- If a substance is kept below its freezing point, it may become more or less dangerous.
- The freezing point additionally offers a crucial safety standard for evaluating the impacts of worker exposure to cold conditions.
Learn moree about Colligative Properties here:
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Answer:
17 protons
19 neutrons
Explanation:
Chlorine will always have the same amount of protons, and that would be 17 protons.
The atomic mass will change according to how many neutrons are present.
Cl - 35 is comprised of 17 protons and 18 neutrons.
We want to find Cl - 36:
We simply add 1 neutron. 18 + 1 = 19 neutrons.
Answer:
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Explanation:
Answer:
Final temperature = 83.1 °C
Explanation:
Given data:
Mass of concrete = 25 g
Specific heat capacity = 0.210 cal/g. °C
Initial temperature = 25°C
Calories gain = 305 cal
Final temperature = ?
Solution:
Q = m. c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C
305 cal = 5.25cal/°C × T2 - 25°C
305 cal / 5.25cal/°C = T2 - 25°C
58.1 °C = T2 - 25°C
T2 = 58.1 °C + 25°C
T2 = 83.1 °C