Answer:
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Explanation:
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:
(a) 33.6 L of oxygen would be produced.
(b) 106 grams of 
would be needed
<h3>Stoichiometric calculations</h3>
1 mole of gas = 22.4 L
(a) From the equation, 2 moles of 
produce 3 moles of 
. 1 mole of will, therefore, produce 1.5 moles of .
1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L
(b) 22.4 L of 
is produced at STP. This means that 1 mole of the gas is produced.
From the equation, 1 mole of requires 1 mole of .
Molar mass of = (23x2)+ (12)+(16x3) = 106 g/mol
Mass of 1 mole = 1 x 106 = 106 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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