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mars1129 [50]
3 years ago
5

Which electrolyte is used in an alkali fuel cell?

Chemistry
2 answers:
zaharov [31]3 years ago
4 0
 i think..
potassium hydroxide..
xxTIMURxx [149]3 years ago
3 0
One major component of all fuel cells is the electrolyte. An electrolyte is a solution that is able to conduct electricity. In an alkaline fuel cell the electrolyte is an alkaline liquid: in this case, potassium hydroxide also known as KOH.
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The solubility of kcl in ethanol is 0.25 g / 100 ml at 25 oc. how does this compare to the solubility of kcl in water?
Furkat [3]
Solubility data of a certain solute with a certain solvent is empirical. There are constant values for this at varying temperatures. For KCl in water at 25°C, the solubility is 35.7 g/100 mL of water. When you compare this with the solubility data of KCl with ethanol, this means that KCl is more soluble in water than in ethanol. This is true because KCl is an ionic salt which is very soluble in water.
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3 years ago
A farmer has a 10 acre plot to feed his family. What is the most efficient way to use the plot?
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6 0
3 years ago
A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base
Gekata [30.6K]
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
6 0
3 years ago
Read 2 more answers
You have a 100 ml stock solution of 100 mg/ml ampicillin in deionized water. You want to make 30 ml of 25 mg/ml ampicillin in de
Georgia [21]

Explanation:

Volume of the stock solution is V_{1} = ?

Initial concentration of ampicillin is M_{1} = 100 mg/ml

Final volume (V_{2}) = 30 ml

Final concentration of  ampicillin (M_{2}) = 25 mg/ml

Therefore, calculate the volume of given stock as follows.

              M_{1} \times V_{1} = M_{2} \times V_{2}

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

            M_{1} \times V_{1} = M_{2} \times V_{2}

            100 mg/ml \times V_{1} = 25 g/ml \times 30 ml    

                   V_{1} = 7.5 ml

Now, we will calculate the volume of water added into it as follows.

            Volume of water added = V_{2} - V_{1}

                                                  = 30 ml - 7.5 ml

                                                  = 22.5 ml

Thus, we can conclude that required solution is 22.5 ml of deionized water.

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3 years ago
I need help ill give points plssss help
irina1246 [14]
A I think because each side cancels out the other opposite side
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