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3 years ago

If 25.16 g of chlorine react with 12.99 g of manganese metal, what is the empirical formula of the compound?

1 answer:

6 0

We use the given masses of the reactants to calculate the moles of Mn and Cl. Empirical formula represents the simplest mole ratio of atoms present in a compound.

Moles of Mn = $12.99 g Mn * \frac{1 mol Mn}{54.94 g Mn} = 0.236 mol Mn$

Moles of Cl = $25.16 g Cl_{2} *\frac{1 mol Cl_{2}}{70.91 g Cl_{2}} * \frac{2 mol Cl}{1 mol Cl_{2}}$ = 0.710 mol Cl

Simplest mole ratio: $Mn_{\frac{0.236}{0.236}}Cl_{\frac{0.710}{0.236}}$

So the empirical formula is $MnCl_{3}$

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3 years ago

Select the properly written and balanced equation for the following chemical reaction: Sodium metal and an aqueous solution of m

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3 years ago

Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0

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When E° cell is an electrochemical cell which comprises of two half cells.

So,

when we have the balanced equation of this half cell :

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and we have also this balanced equation of this half cell :

Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

= 2.46 V

∴ the correct answer is 2.46 V

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3 years ago