It is not good conductors of electricity or heat!
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Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,
Formula used :
We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
![\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]](https://tex.z-dn.net/?f=%5CDelta%20%28H_%7Bf%7D%29_%7Breaction%7D%3D%5B2moles%5Ctimes%20%28-1699.8%20KJ%2Fmole%29%7D%2B3moles%5Ctimes%20%280%5Ctext%7B%20KJ%2Fmole%7D%7D%29%5D-%5B%283moles%5Ctimes%28-520.0KJ%2Fmole%20%7D%2B4moles%5Ctimes%280%5Ctext%7B%20KJ%2Fmole%7D%29%5D)
= (-3399.6) + (1560)
= -1839.6 KJ
Answer:
The specific heat of iron is 0.45 J/g.°C
Explanation:
The amount of heat absorbed by the metal is given by:
heat = m x Sh x ΔT
From the data, we have:
heat = 180.8 J
mass = m = 22.44 g
ΔT = Final temperature - Initial temperature = 39.0°C - 21.1 °C = 17.9°C
Thus, we calculate the specific heat of iron (Sh) as follows:
Sh = heat/(m x ΔT) = (180.8 J)/(22.44 g x 17.9°C) = 0.45 J/g.°C
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
