Answer:
Explanation:
Depression in freezing point is:
where,
= freezing point of solution = ?
= freezing point of solvent (cyclohexane) =
= freezing point constant of solvent (cyclohexane) =
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute (biphenyl) = 0.771 g
= mass of solvent (cyclohexane) = 25.0 g
= molar mass of solute (biphenyl) =
Now put all the given values in the above formula, we get:
Therefore, the freezing point of a solution made by dissolving 0.771 g of biphenyl in 25.0 g of cyclohexane is
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
The main pieces of evidence that an exothermic reaction has occurred is an increase in temperature due to the release of energy, a release of energy in the form of light, or a release of gas.
Answer:
1 gram of H2 will be produced from 12 grams of Mg.
Explanation:
According to Stoichiometry, 0.5 moles of Mg are present. 1 mole of Mg produces 1 mole of H2, so 0.5 moles of Mg will produce 0.5 moles of H2. Multiplying molar mass of H2 i.e. 2 gram/mole with 0.5 moles, we can find the mass of H2 in grams which is 1 gram.