Okay so, 24.00 mL of a 0.25 M NaOH solution is titrated.
Answer: the first option is correct
Why: the seed is the one doing all the work and no other forces were mentioned after 4 days
The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.
Option D
<h3>Chemical Reactions</h3>
Generally,the experiment shows that glass B temperature is higher than glass temperature A and this is given that observation that the solute dissolves faster in glass B than glass A.
Therefore,The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.
For more information on Temperature
brainly.com/question/13439286
Answer:
Higher the frequency, the higher the energy
Explanation:
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Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
