Answer:
The temperature is the same overtime.
Explanation:
Since the line on the graph is straight the temperature will be constant.
Answer:
2.6%
Explanation:
As, 1 ounce (oz) = 0.0625 pounds (lb)
Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb
Since, 1 oz = 0.0625 lb
⇒ 4 oz = 4×0.0625 = 0.25 lb
Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb
The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.
Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>
Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.
Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>
Yes that is true, when energy is transformed energy is not destroyed.
<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.
<u>Explanation:</u>
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:
Where,
= change in Gibb's free energy
= change in enthalpy
T = temperature
= change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and 
is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the is also negative.
For a reaction to be spontaneous, must be negative.
![-ve=-ve-[T(-ve)]\\\\-ve=-ve+T](https://tex.z-dn.net/?f=-ve%3D-ve-%5BT%28-ve%29%5D%5C%5C%5C%5C-ve%3D-ve%2BT)
From above equations, it is visible that will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.
You need to find the abundance. Then, multiply the abundance by 100, and add that to the mass for each isotope. Basically, for each isotope, take the percentage abundance and add it to the mass. Multiply each calculation of these together to get your average atomic mass,
