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SIZIF [17.4K]
3 years ago
11

Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 28.7 moles of FeO. Y

ou must show all of your work. 2Al + 3FeO --> 3Fe +Al2O3
Chemistry
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

                            FeO        :           Al

                             2            :            3

                          28.7          :         3/2×28.7 = 43.05 mol

Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.

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Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b
Alinara [238K]

Answer:

Explanation:

The cell reaction properly written is shown below:

              Cu|Cu²⁺_{aq} || Ag⁺_{aq} | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

  Oxidation half:

                  Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

At the anode, oxidation occurs.

  Reduction half:

                  Ag⁺_{aq} + 2e⁻ ⇄ Ag_{s}

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

             Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

              Ag⁺_{aq} + e⁻ ⇄ Ag_{s}

  we multiply the second reaction by 2 to balance up:

         2Ag⁺_{aq} + 2e⁻ ⇄ 2Ag_{s}

The net reaction equation:

Cu_{s} + 2Ag⁺_{aq} + 2e⁻⇄ Cu²⁺_{aq} + 2e⁻ + 2Ag_{s}

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

  Cu_{s} + 2Ag⁺_{aq} ⇄ Cu²⁺_{aq} + 2Ag_{s}

6 0
2 years ago
Solid added to liquid and the solid floats down to bottom of container physical or chemical change?
Makovka662 [10]

Answer:

Physical change

Explanation:

4 0
3 years ago
3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.
igor_vitrenko [27]
Volume percent = Volume of solute
                              ----------------------------------
                                Volume of the solution
  
                  2                    Volume of the solute
               -------   =           ------------------------------
                100                               250
                     
         Volume of the solute = 2 x 250
                                                ------------
                                                  100         

                                             =   5 mL.

Hope this helps!




                            




4 0
2 years ago
Olivia, a Latina student, is told that she can check only two books out of the library at a time, but Leann, a white student, is
Sliva [168]

Answer:

a

Explanation:

the others are rude, and rather support this, while a helps to support the ending of white privlige

4 0
2 years ago
Read 2 more answers
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
2 years ago
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