Your sense of touch is based in your _____. bones muscles skin tissues

Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) Le Châtelier's principle tells us that an increase in either [A+] or [B−]

Arlecino [84]

Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. $3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

Explanation:

A

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$K_{sp}=S\times (2S)^2=4S^3$

$8.45\times 10^{-12}=4S^3$

Solving for S:

$S=0.000128 M$

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of $M(NO_3)_2$ = 0.202 M

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

So, $[M^{2+}]=0.202 M+S$

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$8.45\times 10^{-12}=(0.202 M+S)(2S)^2$

Solving for S:

$S=3.23\times 10^{-6} M$

$3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

8 0

4 years ago

Calculate the solubility of copper (II) iodate in 0.16 M copper (II) nitrate. Ksp* is 7.4x10-8 M3. *You should know that the Ksp

Simora [160]

Answer:

solubility in presence of 0.16M Cu(IO₃⁻)₂ = 3.4 x 10⁻⁴M*

Explanation:

Cu(IO₃⁻)₂ ⇄ Cu⁺² + 2(IO₃⁻)

C(i) ---------- 0.16M 0M

ΔC ---------- +x +2x

C(f) ---------- 0.16 + x ≅ 0.16M* 2x

Ksp = [Cu⁺²][IO₃⁻]²

7.4 x 10⁻⁸M³ = 0.16M(2x)² = 0.64x²

x = solubility in presence of 0.16M Cu(IO₃⁻)₂ = SqrRt(7.4x10⁻⁸M³/0.64M²)

= 3.4 x 10⁻⁴M*

*Note: This is consistent with the common ion effect in that a reduction in solubility is expected. The normal solubility of Cu(IO₃⁻)₂ in pure water at 25°C is ~2.7 x 10⁻³M.

3 0

3 years ago

P4+O2---->P4 O6 Método algebraico

In-s [12.5K]

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5 0

3 years ago

Calculate Δ H° for the reaction C 4H 4( g) + 2H 2( g) → C 4H 8( g), using the following data: Δ H° combustion for C 4H 4( g) = –

arlik [135]

Answer: The enthalpy of the reaction is coming out to be 2231 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$