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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:
Lets let our mass equal 3 on alletals and solve using d=m/v equation
Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium
Answer:
heat energy is released into the surrounding
Explanation:
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