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Dennis_Churaev [7]
3 years ago
10

How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifre

eze? Use the six-step method.
...?
Chemistry
2 answers:
Sveta_85 [38]3 years ago
7 0
<span>7X+.25*100=.6(100+X)
.7X+25=60+.6X
.7X-.6X=60-25
.1X=35
X=35/.1

X=350 GALLONS OF 70% ANTIFREEZE WILL BE NEEDED.
PROOF
.7*350+25=.6*450
245+25=270
270=270</span>
Anarel [89]3 years ago
6 0
Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons
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<u>Explanation:</u>

The given chemical cell follows:

Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)

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Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.13-(0.0)=0.13V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}

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E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

[H^{+}]=1.00M

[Sn^{2+}]=0.350M

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Putting values in above equation, we get:

E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})

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