Answer:
True
Explanation:
It is true as metals form a positive ion when they take part in a chemical reaction. Nonmetals form an anion when bonding.
:)
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)
Given s = 7.4×10⁻³ M
So, Ksp is:
Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
A sparingly soluble metal fluoride with formula MF₂, where m is an unknown metal, has a Ksp = 6.65 x 10⁻⁶. The concentration of F⁻ in solution 2.36 x 10⁻³M
Ksp is called the Molar solubility product and S is the Molar solubility of an ion in a solution.
According to given formula, the dissociation of metal fluoride MF₂ occurs as follows in aqueous solution:
MF₂ ------> M⁺² + 2F⁻
S 2S
Ksp = [S] [2S]²
Ksp = 4S³
Given, Ksp = 6.65 x 10⁻⁶
On substituting,
6.65 x 10⁻⁶ = 4S³
S³ = 1.66 x 10⁻⁶
S = 1.18 x 10⁻³
So, Solubility of F⁻ ion = 2S
Solubility of F⁻ ion = 2(1.18 x 10⁻³)
Solubility of F⁻ ion = 2.36 x 10⁻³
Since Molarity of the ions is equal to the solubility of the ion in aqueous solution.
Hence, the concentration of F⁻ ion is 2.36 x 10⁻³M.
Learn more about Molar solubility here, brainly.com/question/16243859
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D. Light waves are electromagnetic and sound waves are mechanical waves.
<span>a. Lack of precise measuring instruments.</span>
