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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A atomic number hope i helped
<h3>
Answer:</h3>
70.906 g
<h3>
Explanation:</h3>
We are given;
- Atoms of Chlorine = 1.2 × 10^24 atoms
We are required to calculate the mass of Chlorine
- We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
- That is , 1 mole of an element = 6.022 × 10^23 atoms
- Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms
But since Chlorine gas is a molecule;
- 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms
But, molar mass of Chlorine gas = 70.906 g/mol
Then;
70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms
= 1.20 × 10^24 atoms
Thus;
For 1.2 × 10^24 atoms ;
= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)
<h3>= 70.906 g </h3>
Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g
=
Answer:
5 moles of NO₂ will remain after the reaction is complete
Explanation:
We state the reaction:
3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)
3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:
If 3 moles of nitric oxide need 1 mol of water to react
Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O
We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:
1 mol of water needs 3 moles of oxide to react
Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide
We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains
