Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
1 answer:
Answer:
There was 450.068g of water in the pot.
Explanation:
Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L
Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s
Let m = x g be the weight of water in the pot.
Energy required to vaporise water = mL = 2260x
Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x
Total energy required =
Hence, there was 450.068g of water in the pot.
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First, we need to find the atomic mass of 
.
According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16
As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of = 6 * 12.01 + 12 * 1.008 + 6 * 16
The molar mass of = 180.156 grams/mole
Now that we have the molar mass of, we can find the grams of glucose by using:
mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)
Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams
Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams
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According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16
As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of
The molar mass of
Now that we have the molar mass of
mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)
Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams
Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams
-i