23. The cell has a

Which metal will react spontaneously with Cu2+ (aq) at 25°C?

Gwar [14]

Answer:

Mg

Explanation:

The standard reduction potentials are

E°/V

Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42

Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85

Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34

Mg2+(aq) + 2e- ⟶ Mg(s); -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

5 0

3 years ago

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Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

Answer: The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

Explanation:

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago

Which is the formula for magnesium nitride?

Mekhanik [1.2K]

Explanation:

Mg₃N₂ is the formula of magnesium nitride.

5 0

3 years ago

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Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s

Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0

3 years ago

What is the PEN number for Oxygen?

nataly862011 [7]

Answer:

8

Explanation:

g o o g l e d it

7 0

2 years ago