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3 years ago

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

Chemistry

2 answers:

Gwar [14]3 years ago

5 0

Answer:

Explanation:

The standard reduction potentials are

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⁺.

Oksi-84 [34.3K]3 years ago

3 0

Answer:

B. Mg

Explanation:

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4 years ago

A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point

Ksivusya [100]

Answer:

$P_{T}$ = 2.94 atm

Explanation:

The total pressure ( $P_{T}$ ) in the container is given by:

$P_{T} = P_{O_{2}} + P_{He}$

The pressure of the oxygen ( $P_{O_{2}}$ ) and the pressure of the helium ( $P_{He}$ ) can be calculated using the ideal gas law:

$PV = nRT$

<u>Where</u>:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

$n_{O_{2}} = \frac{m}{M}$

Where m is the mass of the gas and M is the molar mass

$n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles$

And the number of moles of helium is:

$n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles$

Now, we can find the pressure of the oxygen and the pressure of the helium:

$P_{O_{2}} = \frac{nRT}{V} = \frac{1.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 0.98 atm$

$P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm$

Finally, the total pressure in the container is:

$P_{T} = P_{O_{2}} + P_{He} = 0.98 atm + 1.96 atm = 2.94 atm$

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!

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4 years ago