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aliya0001 [1]
3 years ago
12

Jonas and his family are moving to another part of the city. As Jonas, his brother, and his Dad were driving one of the trucks f

illed with furniture to their new home, the truck ran out of gas. A gas station is 0.5 miles down the road. The three of them decide they need to somehow get the truck to the gas station to get gas. Jonas tries pushing the truck, but can not make it move. Which of the following would NOT be a solution to help get the truck to the gas station?
Unload the furniture out of the truck. With less mass, they should be able to push the truck to the gas station.

Jonas and his brother push the truck. With more force, they should be able to push the truck to the gas station.

Jonas’ brother gets out of the cab of the truck and sits in the back of the truck with the furniture. With less mass, they should be able to push the truck to the gas station.

Unload half the furniture out of the truck and let Jonas and his brother push. With less mass and more force, they should be able to push the truck to the gas station.
Physics
2 answers:
Nimfa-mama [501]3 years ago
7 0
Jonas and his brothers push the truck C
vredina [299]3 years ago
5 0

Answer:

i think the answer would be:Jonas’ brother gets out of the cab of the truck and sits in the back of the truck with the furniture. With less mass, they should be able to push the truck to the gas station.

Explanation:

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ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami
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Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

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The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{product}

\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

\Delta S^o=132.67J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

\Delta G^o=22664.34J=22.66kJ

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

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Answer:

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