1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
IrinaVladis [17]
3 years ago
10

HELP PLEASE. 20 POINTS AND BRAINLIEST!!!! Solubility of salts in water is temperature dependent. Consider the solubility curve o

f the salts seen here. Which salt's solubility changed the least from 0 - 100oC; which salt's solubility changed the greatest amount from 0 - 100oC?

Chemistry
2 answers:
Zina [86]3 years ago
6 0

Answer: Sodium chloride's solubility has changed the least from  0^0C-100^0C and potassium nitrate's solubility has changed the most from 0^0C-100^0C.

Explanation:

Solubility in a particular liquid is the amount of solute that can dissolve in unit volume of the liquid to form the saturated solution at the given temperature and under a pressure of 1 atmosphere.

a) Solubility of Potassium iodide at 0^0C: 129g/100 ml

Solubility of Potassium iodide at 100^0C: 250g/10ml

Change in solubility of Potassium iodide = (250-129)^0C=121^0C

b) Solubility of sodium nitrate at 0^0C: 70g/100 ml

Solubility of sodium nitrate at 100^0C: 180 g/100 ml

Change in solubility of sodium nitrate  = (180-70)^0C=110^0C

c) Solubility of Potassium nitrate at 0^0C: 12g/100 ml

Solubility of Potassium nitrate at 100^0C: 235g/100 ml

Change in solubility of Potassium nitrate = (235-12)^0C=223^0C

d) Solubility of sodium chloride at 0^0C: 36g/100 ml

Solubility of sodium chloride at 100^0C: 40 g/100 ml

Change in solubility of sodium chloride= (40-36)^0C=4^0C

e) Solubility of Potassium chlorate at 0^0C: 5g/100 ml

Solubility of Potassium chlorate at 100^0C: 60 g/100 ml

Change in solubility of Potassium chlorate = (60-5)^0C=55^0C

Thus we can see that sodium chloride's solubility has changed the least from  0^0C-100^0C and potassium nitrate's solubility has changed the most from 0^0C-100^0C.

Dennis_Churaev [7]3 years ago
4 0
Sodium chloride's solubility only changed about 5 g/100 mL water, whereas potassium nitrate's solubility changed about 230 g/100 mL water
You might be interested in
How many millimeters are there in 1 kilometer
Marysya12 [62]
1.000.000 is the correct answer
4 0
3 years ago
Consider the unbalanced equation for the oxidation of aluminum.
SashulF [63]
In order to balance an equation, we apply the principle of conservation of mass, which states that mass can neither be created nor destroyed. Therefore, the mass of an element before and after a reaction remains constant. Here, the balanced equation becomes:
4Al + 3O₂ → 2Al₂O₃

The coefficients are 4, 3 and 2.
8 0
3 years ago
Read 2 more answers
What is true of a basic solution at room temperature? it has a ph value below 7. it has a greater concentration of hydroxide com
slega [8]
The true statement about basic solution at room temperature is that it has a greater concentration of hydroxide compared to hydronium ions.
Basic solutions have always pH greater than 7.
Basic solutions have bitter and caustic taste.
Basic solutions are not used as conductors in car batteries, acidic electrolytes are used in car batteries.
7 0
3 years ago
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero d
Yuki888 [10]
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g 
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK 
</span>
Now, energy required for melting of ICE = <span>  334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span>  2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2)  + (3)
                                                                        = 1753.5 + 2195.18 + 11849.25
                                                                        = </span><span>15797.93 J 
</span><span>                                                                        = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
7 0
3 years ago
How many moles of ions would you expect in an aqueous solution containing one mole of chromium(III) chloride? Hint: write out th
Cerrena [4.2K]

Answer:

Four  

Explanation:

AlCl₃(aq) ⟶ Al³⁺(aq) + 3Cl⁻(aq)

One mole of AlCl₃  produces 1 mol of Al³⁺ and 3 mol of Cl⁻.

That's four moles of ions.

8 0
3 years ago
Other questions:
  • Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with
    13·1 answer
  • The coefficients in a balanced chemical equation represent
    8·1 answer
  • If a 0.75 mole sample of helium gas occupies a volume of 1.5 l, what volume will a 1.2 mole sample of gas occupy at the same tem
    6·1 answer
  • Name the following compound: <br> HC2H3O2
    11·2 answers
  • How many carbons are in a molecule of hexane?<br> a. 2<br> b. 4<br> c. 6<br> d. 8
    14·1 answer
  • !DUE SOON PLEASE HELP!
    7·1 answer
  • Highlight the basic points in Lewis and Langmuir theory of electrovalency
    6·1 answer
  • The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
    7·1 answer
  • G Calculate the mass, in grams, of 225.0 atoms of cadmium, Cd (1 mol of Cd has a mass of 112.41 g).
    7·1 answer
  • Calculate 28.1g of silicon to moles
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!