Answer:
0.366m = Molality of the solution
Explanation:
To solve this question we must know the addition of a solute produce decreasing in freezing point regard to the pure solvent. The equation is:
ΔT = m*Kf*i
<em>Where ΔT is change in freezing point </em>
(As freezing point of water is 0°C, the ΔT is 2.72°C)
<em>Kf is freezing point depression constant = 1.86°C/m for water</em>
<em>i is Van't Hoff factor. The number of ions produced when 1 mole of the salt is dissolved = 4 ions for Fe(ClO₃)₃, Fe³⁺ and 3 ClO₃⁻ ions</em>
<em>m is molality of the solution.</em>
<em />
Replacing:
2.72°C = m*1.86°C/m*4
<h3>0.366m = Molality of the solution</h3>
<em />
Endothermic means it absorb energy from surrounding and its sign is positive so when it absorb this energy the flask or the container become colder
Answer:
PubChem CID 6393
Structure
Find Similar Structures
Chemical Safety
Compressed Gas
Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula CF4
Molecular Weight
88.004 g/mol
Explanation:
1) An endothermic reaction is a reaction where heat is "absorbed" and the surrounding area is cooled. An example of an endothermic reaction is the classic baking soda and vinegar reaction. The mixture gets cold.
An exothermic reaction is one which gives off heat, such as burning, which is itself a chemical reaction.
2) (I don't actually know this one but i hope i helped with number one)
There are no numbers provided so I will try to explain the equation. If you want to find the percent error you would do theoretical (the number it’s supposed to be) minus experimental (the number you got in the experiment) over the theoretical value (same as the first one.) You must multiply this by 100 in order to show the actual percentage rather than very low decimals. You want the percent error to be as low as possible because the lower it is the less mistakes happened. Mistakes can include too much/less of solvent, solution, or the glassware wasn’t clean enough, and much much more.
I hope this helps!
Please mark the brainliest if it does! (: