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2 years ago

how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution

1 answer:

4 0

0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.

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3 years ago

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2 years ago

We could avoid a large increase in temperature if greenhouse emissions peaked by the year _____.

Naya [18.7K]

Answer:

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Explanation:

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3 years ago

Consider the concentration cell in which the metal ion has a charge of +2, and the solution concentrations are: dilute solution

tia_tia [17]

Answer: The predicted cell potential of the cell is +0.0587 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $M(s)\rightarrow M^{2+}+2e^-$

Reduction half reaction (cathode): $M^{2+}+2e^-\rightarrow M(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[M^{2+}_{(diluted)}]}{[M^{2+}_{(concentrated)}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[M^{2+}_{(diluted)}]$ = 0.05 M

$[Zn^{2+}_{(concentrated)}]$ = 4.808 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{0.05M}{4.808M}$

$E_{cell}=0.0587V$

Hence, the predicted cell potential of the cell is +0.0587 V

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2 years ago

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Alexandra [31]

Answer:

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