Chemistry, 16.11.2020 17:30 batmanmarie2004

10g of Compound X with molecular formula are burned in a constant-pressure calorimeter containing 45g of water at 25. The temperature of the water is observed to rise by 2.432. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound at .

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Answer from: flore9584

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10 g of Compound X with molecular formula C_4 H_8 are burned in a constant-pressure calorimeter containing 45g of water at 25^oC . The temperature of the water is observed to rise by 2.432. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound at 25^oC

The value is $\Delta H^o_f =-2571 \ J /mol$

Explanation:

From the question we are told that

The mass of compound X is $m = 10 \ g = 0.010 \ kg$

The mass of water is $m_w = 45 \ g = 0.045 \ kg$

The temperature of water is T_w = 25^oC

The change in the temperature of water is $\Delta T = 2.432 ^oC$

Generally the heat adsorbed by water is mathematically represented as

$Q_{w } = c * m _w * \Delta T$

Here c is the specific heat of water with value $c = 4186 \ J/kg\cdot ^oC$

$Q_{w } = 4186 * 0.045 * 2.432$

=> $Q_{w } = 458.1 \ J$

Given that the total heat that was generated by the reaction is absorbed by water then

$\Delta H _{rxn} = -Q_w$

The negative sign shows that the heat was absorbed

$\Delta H _{rxn} = -458.1 \ J$

Generally the number of moles of the compound X available is mathematically represented as

$n = \frac{m}{Z}$

Here Z is the molar mass of compound X the value is $Z = 56.11 \ g/mol$

$n = \frac{m}{Z}$

=> $n = \frac{10}{56.11}$

=> $n = 0.1782 \ mols$

Generally the standard heat of formation of Compound X is mathematically represented as

$\Delta H^o_f = \frac{\Delta H_{rxn} }{ n}$

=> $\Delta H^o_f = \frac{-458.1 }{ 0.1782}$

=> $\Delta H^o_f =-2571 \ J /mol$

Answer from: Quest

Mass is the amount of matter in an object. weight is the measurement of the pull of gravity on an object. say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165) now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon. however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant. in conclusion it is possible for your weight to change as your mass remains constant.

Answer from: Quest

Change of state has wrong energy change - option 2

brian made this table to organize his notes on changes of state. depostion is a change of state that has the wrong energy change listed. it is utilized to collect data as a portion of the development means and in restricted conditions which may be applied at test.

a deposition in the authority of the united states or research for development in the authority of canada includes the catching of affirmed. the out-of-court verbal assertion of a spectator that may be lessened to a printed record for succeeding application for discovery purposes.

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