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larisa86 [58]
3 years ago
5

What element is represented by the chemical symbol V​

Chemistry
2 answers:
arlik [135]3 years ago
7 0
The element represented by V is Vanadium.
NemiM [27]3 years ago
4 0

Answer:

Vanadium

Explanation:

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Explanation: Alcohol floats on oil and water sinks in oil. Water, alcohol, and oil layer well because of their densities, but also because the oil layer does not dissolve in either liquid. The oil keeps the water and alcohol separated so that they do not dissolve in one another. ... Water sinks because it is more dense than oil.

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Which of the following statements about decomposition reactions is true?
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Match each term with its definition. 1) reaction coordinate
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3 years ago
Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra
maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4  mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction of Mg(OH)2 can be given as:

 

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

 Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²  

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

 X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+  = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X *(0.130)²  

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0
3 years ago
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