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ANTONII [103]
3 years ago
7

When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2

.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Use 3 – 4 sentences to explain your answer.
Please Help!
Will give most brainly
Chemistry
2 answers:
Mademuasel [1]3 years ago
6 0

Answer is: boiling point will be changed by 4°C.

Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).

Change in boiling point: ΔT =i · Kb · b.

Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.

b -  molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.

i - Van't Hoff factor.

Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.

Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.

nlexa [21]3 years ago
4 0

The answer should be 4°C....

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3 years ago
3. What is SI system? Why has SI system been developed? Give reasons​
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8 0
2 years ago
Read 2 more answers
Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k
bulgar [2K]

Answer:

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

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 = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

8 0
2 years ago
I don't know how to do this, pls help!
Aliun [14]
Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)

1) LiF smaller  cation then KF
1,036                                 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
</span><span>I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.

2) The same cation Li , so wrong statement.

3)</span>The same cation Na , so wrong statement.

4) NaCl smaller cation then KF
  786                                   853
7 0
2 years ago
Read 2 more answers
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