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3 years ago

What is the mass of 6.00 cm^3 of mercury, density = 13.5939 g/cm^3?

Chemistry

1 answer:

Molodets [167]3 years ago

5 0

Answer: 81.5634kg

Explanation:

Density=mass/volume

13.5939=m/6

make m subject of formula

m= 13.5939×6

m=81.5634kg

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We know,

$\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol$

For given reaction, $I_2(g)+Cl_2(g)\ -->\ 2ICl(g)$

$\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol$

For , 2.41 moles of $I_2$ :

$\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ$

We know :

$\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K$

Hence, this is the required solution.

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2 years ago

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3 years ago

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$Al^0-3e^----\ \textgreater \ Al^{3+}$

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

$Mn^{2+}+2e^{-}---\ \textgreater \ Mn$

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

$2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

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The right side has 2 Al, 3 Mn, and 2*3 positive charges.

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As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.

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