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olga2289 [7]
3 years ago
9

Archimedes himself had a mass of 52 kg when he discovered displacement. If his density was roughly 12 g/mL, calculate the amount

of water he would spill out of a full tub.
Chemistry
2 answers:
trapecia [35]3 years ago
8 0

"He who lays up treasures on earth spends his life backing away from his treasures. To him, death is loss. He who lays up treasures in heaven looks forward to eternity; he’s moving daily toward his treasures. To him, death is gain." - Randy Alcorn

jeka57 [31]3 years ago
8 0

Answer:

7.75 kg

Explanation:

Archimedes’ principle tells you that the weight of the water displaced is equal to the buoyancy force:

To keep the wood afloat, the buoyancy force must have the same magnitude as the force of gravity on the block, so

The volume of water displaced is

So the mass of water displaced is

Thus, the mass of the piece of wood is 7.75 kilograms.

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<u>Answer:</u> The \Delta G for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=-803kJ=-803000J      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=-4.05J/K

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J

For the given chemical equation:

CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)

The expression for K_c is given as:

K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}

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[O_2]=9.27M

Putting values in above equation, we get:

K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}

K_c=0.126

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_c

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = 8.314J/K mol

T = Temperature = 293 K

K_c = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))

\Delta G=-806859.46J=-806.86kJ

Hence, the \Delta G for the reaction is -806.86 kJ

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