Answer:Predatorrrrrrrrrrrrrrrr
Pls gimme Brainliest
Solution :
a). Applying the energy balance,



![$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$](https://tex.z-dn.net/?f=%240%3D%5Bmc%28T_f-T_i%29_%7Biron%7D%5D%20%2B%20%5Bmc%28T_f-T_i%29_%7Bwater%7D%5D%24)


b). The entropy change of iron.


= -9.09 kJ-K
Entropy change of water :


= 10.76 kJ-K
So, the total entropy change during the process is :

= -9.09 + 10.76
= 1.67 kJ-K
c). Exergy of the combined system at initial state,






Therefore, energy of the combined system at the initial state is

= 63.94 -13.22
= 50.72 kJ
Similarly, Exergy of the combined system at initial state,





Thus, energy or the combined system at the final state is :

= 216.39 - 9677.95
= -9461.56 kJ
d). The wasted work



= 50.72 + 9461.56
= 9512.22 kJ
Answer:
2KClO3 》》2KCl +3O2
C+ O2》》CO2
number of C moles
Required O2 moles (According to the mole ratio )
Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles
Now you can find the mass of KClO3
I mentioned the useful steps which can guide you to get the answer.
Explanation:
Answer:
4.2 x 10⁴ mL
Explanation:
Data Given:
Density (d) of air = 1.19 x 10⁻³g/mL
Mass of the air (m) = 50 g
Volume of the air (V) = ?
Solution:
Formula will be used
d = m/V
As we have to find volume so rearrange the above equation
V = m/d . . . . . . . . . . . (1)
Put values in above equation 1
V = 50 g / 1.19 x 10⁻³g/mL
V = 4.2 x 10⁴ mL
So,
volume of dry air = 4.2 x 10⁴ mL