Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
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Answer:
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Explanation:
Iodine is decolorized.
The first reaction stated in the question occurs as follows;
2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)
The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.
Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.
The equation of the titration reaction is;
2Na2S2O3 + I2→ Na2S4O6 + 2NaI
When this reaction takes place, iodine is decolorized due to its reduction to I^-.
