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kobusy [5.1K]
4 years ago
10

Unless the gasoline tank of a car is refilled, the car will stop moving when it runs out of gas. What happened to the energy tha

t was stored in the gasoline? a. It was eventually destroyed. b. It was converted into matter. c. It was transfered to the roadway. d. It was converted to mechanical energy and heat.
Chemistry
1 answer:
GaryK [48]4 years ago
3 0

Answer: The correct answer is option d.

Explanation: Energy stored into gasoline is converted into another form of energy. This was very well explained on the basis of Law of conservation of energy. This law states that energy is conserved in the system which means that energy is converted from one form to another form.

The energy stored into gasoline is converted into mechanical energy which helps the car to move and some of the energy is lost in the form of heat.

Hence, the correct answer is Option d.

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Which of the following indicates a reaction with a positive ΔG? A. endergonic, spontaneous B. endergonic, not spontaneous C. exe
shtirl [24]

Answer: B. endergonic, not spontaneous

Explanation:

Endergonic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and  \Delta H for the reaction comes out to be positive.

The Gibbs equation is:

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

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4 0
3 years ago
A 4.00 g sample of a metal (specific heat = 0.600 J g-1°C-1 is heated to 75 degrees Celcius and then dropped into 165 g of water
Bond [772]

Answer:

28.16 °C

Explanation:

Considering that:-

Heat gain by water = Heat lost by metal

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)

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Mass = 165 g

Initial temperature = 28 °C

Specific heat of water = 4.184 J/g°C

For metal:

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Initial temperature = 75 °C

Specific heat of water = 0.600 J/g°C

So,  

165\times 4.184\times (T_f-28)=4.00\times 0.600\times (75-T_f)

690360\left(T_f-28\right)=2400\left(75-T_f\right)

692760T_f=19510080

T_f = 28.16\ ^0C

<u>Hence, the final temperature is 28.16 °C</u>

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