Answer: 15062.4 Joules
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of food = 200.0g
C = 4.184 j/g°C
Φ = (Final temperature - Initial temperature)
= 83.0°C - 65.0°C = 18°C
Then, Q = MCΦ
Q = 200.0g x 4.184 j/g°C x 18°C
Q = 15062.4 J
Thus, 15062.4 joules of heat energy was contained in the food.
Hey mate!
Taxonomy is the <span>study of classification of organisms. Therefore, your answer is A.
Hope this helps!</span>
Lithosphere, asthenosfhere , lower mantle
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
<span>0.38
You first calculate the total moles by dividing the grams by molecular weight:
45 g N2 / 28.02 g/mol = 1.6 mol N2
40 g Ar / 39.95 g/mol = 1.0 mol
Then you divide the moles of Ar by the total number of moles:
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>
