B is right yuh yuh yuh yuh
From the combustion of octane, the formaldehyde will be formed as this equation:
C8H18 + O2 → CH2O + H2O this is the original equation but it is not a balanced equation, so let's start to balance it:
the equation to be balanced so the number of atoms on the right side of the equation sholud be equal with the number of atoms on the lef side.
-we have 8 C atoms on left side and 1 atom on the right side so we will try putting 8 CH2O on the right side instead of CH2O
C8H18 + O2 → 8 CH2O + H2O
we have 2 O atoms on the left side and 9 atoms on the right side so we will try first to put 9 O2 instead of O2 on the left side and put 2H2O on the right side and put 16 CH2O instead of 8 CH2O to make the atoms of O are equal on both sides = 18 atoms
C8H18 + 9 O2 → 16 CH2O + 2H2O
put now we have 8 atom C on the left side and 16 atom on the right side so, we will put 2 C8H18 instead of C8H18 now we get this equation:
2C8H18 + 9O2 →16 CH2O + 2H2O
-now we have 36 of H atoms on both sides.
- and 16 of C atoms on both sides.
- and 18 of O atoms on both sides.
now all the number of atoms of O & C & H are equal on both sides
∴ 2C8H18 + 9O2 → 16 CH2O + 2 H2O
is the final balanced equation for the formation of formaldehayde
Answer:
The answer is definitely D
Explanation:
Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
