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3 years ago

15

which of the following is not a valid conversion factor? A.) 60seconds/1minute B.) 12hours/1day C.) 1dozen/12 cookies D.) 1yard/

3feet

1 answer:

olga2289 [7]3 years ago

6 0

I think the answer is b

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A solution has pOH 5.9

QveST [7]

Answer:

The answer to your question is below

Explanation:

Data

pOH = 5.9

a) [OH⁻] = ?

pOH measures the [OH⁻]

Formula

pOH = -log[OH⁻]

-Substitution

5.9 = -log[OH⁻]

[OH⁻] = antilog (-5.9)

-Result

[OH⁻] = 1.26 x 10⁻⁶ M

b) [H₃O⁺]

pH + pOH = 14

-Solve for pH

pH = 14 - pOH

-Substitution

pH = 14 - 5.9

-Result

pH = 8.1

-Calculate [H₃O⁺]

pH = -log[H₃O⁺]

-Substitution

8.1 = -log[H₃O⁺]

[H₃O⁺] = antilog(-8.1)

-Result

[H₃O⁺] = 7.9 x 10⁻⁹ M

c) This solution is alkaline because the pH is higher than 7.

6 0

3 years ago

Read 2 more answers

The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

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