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Verizon [17]
3 years ago
6

If a student measured an object's density to be 2.70 g/mL, 2.32 g/mL, 2.12 g/mL and 3.88, then his results would considered ____

__The object's actual density is 3.50 g/mL A. Accurate and precise B.Precise but not accurate C. Neither accurate nor precise D.Accurate but precise
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

C. Neither accurate nor precise

Explanation:

  • The student's results would be considered precise if they were close to one another. However they vary significantly from one another.
  • Regarding accuracy, they would be considered accurate if they were close to the actual value. The given results are too far away from the actual value.

Thus the results are neither accurate nor precise.

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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
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Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

4 0
4 years ago
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