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Ksenya-84 [330]
2 years ago
14

The lead-containing reactant(s) consumed during recharging of a lead-acid battery is/are ________. The lead-containing reactant(

s) consumed during recharging of a lead-acid battery is/are ________. PbSO4 (s) only Pb (s) only PbO2 (s) only both PbO2 (s) and PbSO4 (s) both Pb (s) and PbO2 (s)
Chemistry
2 answers:
irga5000 [103]2 years ago
4 0

Answer:  The  lead-containing reactant(s) consumed during recharging of a lead-acid battery is PbSO_4(s)

Explanation:

In lead acid battery, the anode is made up of lead and undergoes oxidation during discharging and cathode is made up of lead oxide and acts as cathode during discharging. The electrolyte used is dilute H_2SO_4.

Charging:

Cathode : reduction : PbSO_4(s)+2e^{-}\rightarrow Pb+SO_4^{2-}

Anode: oxidation : PbSO_4+2H_2O\rightarrow  PbO_2+SO_4^{2-}+4H^++2e^-

Overall reaction : 2PbSO_4+2H_2O\rightarrow Pb+PbO_2+2H_2SO_4

The  lead-containing reactant(s) consumed during recharging of a lead-acid battery is PbSO_4(s)

Mariulka [41]2 years ago
4 0

Answer:

PbSO4 (s) only

Explanation:

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How many liters are in a .00813M solution that contains 1.55 g of KBr?
Sloan [31]

Answer:

                      1.602 L (or) 1602 mL

Explanation:

             Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Rearranging above equation for volume,

                         Volume of solution  =  Moles / Molarity    -------(2)

Data Given;

                  Molarity  =  0.00813 mol.L⁻¹

                  Mass  =  1.55 g

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  1.55 g / 119.002 g.mol⁻¹

                   Moles  =  0.0130 mol

Now, putting value of Moles and Molarity in eq. 2,

                         Volume of solution  =  0.0130 mol / 0.00813 mol.L⁻¹

                         Volume of solution  = 1.60 L

or,

                         Volume of solution  =  1602 mL

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3 years ago
What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

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5 0
3 years ago
Determine the mass of MgCl2 needed to create a 100. ml solution with a concentration of 3.00 M.
kiruha [24]

Explanation:

1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\  = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1}  \: g \\  = 28.5 \: g \: of \: magnesium \: chloride

5 0
2 years ago
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