Answer : The amount of heat released, 45.89 KJ
Solution :
Process involved in the calculation of heat released :

Now we have to calculate the amount of heat released.
where,
Q = amount of heat released = ?
m = mass of water = 27 g
= specific heat of liquid water = 4.184 J/gk
= specific heat of solid water = 2.093 J/gk
= enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole
conversion :
Now put all the given values in the above expression, we get
(1 KJ = 1000 J)
Therefore, the amount of heat released, 45.89 KJ
4KO₂ + 2CO₂ → 2K₂CO₃ + 3 O₂
The two chemicals that have a molar ratio of 3:4 is in equation above O₂ and KO₂
<em><u>explanation</u></em>
The molar ratio is the ratio between the amount in moles of two compounds that are involved in a chemical reaction.
In the equation above the mole ratio is determined by examining coefficient ( <em>number in front of the formula</em>)
the coefficient of O₂ is 3 while that of KO₂ is 4 therefore the mole ratio of O₂:KO₂ is 3:4
Answer:
The concentration of the standard NaOH solution is 0.094 moles/L.
Explanation:
In the titration, the equivalence point is defined as the point where the moles of NaOH (the titrant) and KHP (the analyte) are equal:
moles of NaOH = moles of KHP
![[NaOH]xV_{NaOH} = moles of KHP](https://tex.z-dn.net/?f=%5BNaOH%5DxV_%7BNaOH%7D%20%3D%20moles%20of%20KHP)
![[NaOH] = \frac{moles of KHP}{V_{NaOH}}](https://tex.z-dn.net/?f=%5BNaOH%5D%20%3D%20%5Cfrac%7Bmoles%20of%20KHP%7D%7BV_%7BNaOH%7D%7D)
The
is 40.82mL = 0.04082L and the moles of KHP are

Replacing at the first equation:
![[NaOH] = \frac{3.827x10^{-3}moles}{0.04082L} = 0.094 moles/L](https://tex.z-dn.net/?f=%5BNaOH%5D%20%3D%20%5Cfrac%7B3.827x10%5E%7B-3%7Dmoles%7D%7B0.04082L%7D%20%3D%200.094%20moles%2FL)