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2 years ago

As the tempreture of a liquid increases the solubility of a liquid of that liquid

1 answer:

8 0

As the temperature increases, the solubility of the solute in the liquid also increases. This is due to the fact that the increase in energy allows the liquid to more effectively break up the solute. The additoin of energy also shifts the equilibrium of the reation to the right since it takes energy to dissolve most things and you are adding more of it (this is explained with Le Chatlier principles).

I hope this helps and also I assumed that your question involved the solubility of an ionic substance in a solvent like water. If that was not your question feel free to say so in the comments so that I can answer your actually question.

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The amount of heat absorbed when 5.5g of aluminum is heated from 25.0c to 95.0c the specific heat of aluminum is 0.897J/(gc)

NemiM [27]

Answer:

Q=21.99 kJ

Explanation:

4 0

2 years ago

Which of the following are examples of physical change? Copper oxidizing Iron rusting Carbon dioxide sublimating Water evaporati

kari74 [83]

Carbon dioxide sublimation
water evaporate

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6 0

2 years ago

In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)

jeyben [28]

Answer:

Li2S> Na2S> K2S> CsS

Explanation:

The lattice energy of ionic species depends on the relative sizes of ions in the ionic compounds. As the size of ions increases, the lattice energy decreases and vice versa.

When the size of the anions are the same, the lattice energy now depends on the relative sizes of the cations. Therefore, since all the compounds are sulphides and the order of magnitude of ionic sizes is: Li^+ < Na^+ < K^+ < Cs^+.

Therefore, the order of decrease in lattice energy is; Li2S> Na2S> K2S> CsS

7 0

2 years ago

TRUE or FALSE: The rock closest to the ridge is newer and younger than the older rock

alina1380 [7]

Answer:true

Explanation:

7 0

2 years ago

Calculate the molarity of sodium ion in a solution made

Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

$NaCl\rightarrow Na^{+}+Cl^-$

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

<u> The final concentration of sodium anion = 0.1035 M</u>

4 0

2 years ago