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Which statement correctly describes electrochemical cells?

hoa [83]

Answer: C. Electrochemical cells involve oxidation-reduction reactions.

Explanation: Oxidation occurs at the anode, and reduction occurs at the cathode.

3 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solut

mrs_skeptik [129]

Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

Explanation:

To calculate the number of moles, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

<u>Moles of Aluminium:</u>

Molar mass of aluminium = 27 g/mol

Given mass of aluminium = 0.25g

Putting values in above equation, we get:

$Moles=\frac{0.25g}{27g/mol}=0.00925mol$

<u>Moles of Copper chloride:</u>

Molar mass of copper chloride = 134.45 g/mol

Given mass of copper chloride= 0.40g

Putting values in above equation, we get:

$Moles=\frac{0.40g}{134.45g/mol}=0.00297mol$

For the given chemical equation:

$4Al(s)+3CuCl_2(aq.)\rightarrow 2Al_2Cl_3(aq.)++3Cu(s)$

By Stoichiometry,

3 moles of Copper chloride reacts with 4 moles of aluminium

So, 0.00297 moles of copper chloride will react with = $\frac{4}{3}\times 0.00297$ = 0.00396 moles of Aluminium.

As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.

Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of copper chloride produces 3 moles of copper metal.

So, 0.00297 moles of copper chloride will produce = $\frac{3}{3}\times 0.00297$ = 0.00297 moles of copper metal.

Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:

Molar mass of copper = 63.5 g/mol

Putting values in that equation, we get:

$0.00297mol=\frac{\text{given mass}}{63.5g/mol}$

Given mas of copper metal = 0.188grams (approx 0.20grams)

For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.

Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

6 0

3 years ago

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The heat of vaporization of water at the normal boiling point, 373.2 K, is 40.66 kJ/mol. The molar heat capacity of liquid water

Tatiana [17]

Answer:

$\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, according to the Kirchhoff's law for the enthalpy change, it is possible to compute the heat of vaporization at 300.2 K by considering the following thermodynamic route:

$\Delta _{vap}H(300.2K)=Cp_{liq}(T_b-T\°)+\Delta _{vap}H\°+Cp_{vap}(T-T_B)$

Whereas the first term stands for the effect of taking the liquid from 298.15 K to 373.15 K, the second term stands for the standard enthalpy of vaporization and the last term that of the vapor from the boiling point to 300.2 K; thus we plug in to obtain:

$\Delta _{vap}H(300.2K)=75.37\frac{J}{mol*K} (373.2K-298.15K)+40,660\frac{J}{mol} +36.4\frac{J}{mol*K}(300.2K-373.2K)\\\\\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}$

Best regards!

6 0

3 years ago

Which constants would most likely be filled in the first row of the table?

Triss [41]

Answer:

A. is the answer.

Explanation:

7 0

3 years ago

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1. Explain the two processes that keep atmospheric oxygen and carbon dioxide at stable levels?

lesya [120]

Cellular Respiration and Photosynthesis. Photosynthesis is the process of when plants use sunlight to make foods from carbon dioxide and water, to later on make oxygen. Cellular respiration is the process through which cells convert sugars into energy.

4 0

3 years ago