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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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Degger [83]
<h3>Answer:</h3>

B. C7H16 + 11O2 → 7CO2 + 8H2O

<h3>Explanation:</h3>
  • In a balanced chemical equation, the number of atoms of each element is equal on both sides of the equation.
  • In this case, the balanced chemical equation is;

   C7H16 + 11O2 → 7CO2 + 8H2O

Because, it has 7 carbon atoms, 16 hydrogen atoms and 22 oxygen atoms on each side of the equation.

  • When an equation is balanced it obeys the law of conservation of mass such that the mass of reactants will be equal to the mass of products.
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3 years ago
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
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b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
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c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
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when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
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[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


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3 years ago
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NISA [10]

Answer : The correct option is, Zn^{2+}

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Or we can say that, oxidation reaction occurs when a reactant losses electrons in the reaction.

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Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.

According to the electrochemical series, Zn^{2+} most likely to be reduced because

Hence, the ion most likely to be reduced is Zn^{2+}.

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