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pshichka [43]
3 years ago
13

How are rechargeable batteries recharged? a. Adding new reactants c. Passing electric charge through the reactants b. Changing t

o terminals d. Changing out the electrolytes
Chemistry
1 answer:
olchik [2.2K]3 years ago
4 0
<span>c. Passing electric charge through the reactants Is the answer to you're question.

</span>
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What happens to the individual molecules of a liquid as they gain enough kinetic energy to escape the surface of the liquid?
Bumek [7]
Im pretty sure They freeze
7 0
3 years ago
Read 2 more answers
What is the oh- in a solution with a poh of 5.71
Rudik [331]

Answer:- The hydroxide ion concentration of the solution is 1.95*10^-^6 .

Solution:- The formula used to calculate pOH from hydroxide ion is:

pOH=-log[OH^-]

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

[OH^-]=10^-^p^O^H

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

[OH^-]=10^-^5^.^7^1

[OH^-] = 0.00000195 or 1.95*10^-^6

So, the hydroxide ion concentration of the solution is 1.95*10^-^6 .



3 0
3 years ago
Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0
3 years ago
Plz answer plzplzpzlpl​
astraxan [27]
Whats the question tho?
8 0
3 years ago
Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
olganol [36]

 The  amount  of  Al2O3  in moles=  1.11 moles    while in  grams   = 113.22 grams


    <em><u>calculation</u></em>

     2 Al  + Fe2O3 → 2Fe  + Al2O3

    step  1: find the moles of Al  by  use of <u><em>moles= mass/molar  mass  </em></u>formula

    =  60.0/27= 2.22  moles


    Step 2: use the mole ratio to determine the  moles of Al2O3.

 The  mole ratio  of Al : Al2O3 is  2: 1 therefore the moles of Al2O3= 2.22/2=1.11  moles


Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

7 0
3 years ago
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