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sleet_krkn [62]
2 years ago
7

In this experiment you used an excess of the BaCl2 solution. How would your results be affected if you did not use an excess of

the BaCl2 solution? Would this error cause your calculation of the mass percent of sulfate in the unknown to be too high or too low? Explain.
Chemistry
1 answer:
Igoryamba2 years ago
3 0

if we did not use an excess of the BaCl2 solution it would decrease the mass percentage of sulfate in the unknown sample.

The net precipitation equation would be.

Ba2+(aq) + SO42-(aq)  → BaSO4(s)

If BaCl2 (Ba2+) is not taken in excess then the precipitation would not be completed as some of the sulfate ions would still be remaining in the solution. This would decrease the mass percentage of sulfate in the unknown sample.

If some tiny pieces of filter paper still remained mixed with the precipitate(BaSO4) then the mass of sulfate would increase and it gives a high mass percentage of the sulfate.

mass percentage of sulfate = (mass of sulfate/mass of sample)*100

Learn more about precipitation here brainly.com/question/14675507

#SPJ4

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Answer:

13.A

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Explanation:

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QUESTION 10
iris [78.8K]

Answer:

B

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Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.
sammy [17]
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    </span>S^{2-}/S //  Ni^{2+}/Ni

Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
and {E^{0} S/ S^{2-}  = -0.47 v

Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;
    Cr/Cr^{2+} // Ag^{+}/Ag

Hence, E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}
we know that, {E^{0} Ag^{+}/Ag = -0.22 v
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Therefore, E^{0} cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
8 0
3 years ago
Read 2 more answers
4 Al + 3 O2 --&gt; 2 Al2O3
SIZIF [17.4K]

Answer:

The answer to your question is 7.4 moles of Aluminum

Explanation:

Data

moles of Al = ?

moles of Al₂O₃ = 3.7

Balanced chemical reaction

                4 Al  +  3 O₂  ⇒  2 Al₂O₃

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.

                4 moles of Aluminum ----------------- 2 moles of Al₂O₃

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                            x = (3.7 x 4) / 2

                            x = 14.8 / 2

                            x = 7.4 moles of Aluminum

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What type of mixture will not allow light to pass through
igor_vitrenko [27]
Suspension.  The particles are big enough for the eye to see, and will separate if left sitting.
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