Answer:
All of these stoichiometry problems are worked alike. Here is the 4 step process.
4Na + O2 → 2Na2O
Step 1. Write and balance the equation. You have that.
Step 2. Convert whatever you have to mols. If a solid then mols = grams/molar mass = ?. If a solution then mols = M x L = ? In this case, the molar mass of Na2O is approx 62 so mols = 2/62 = approx 0.03 but that's just a close guess. You should do it more accurately.
Step 3. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want. You have mols Na2O and you want to convert that to mols Na.
0.03 mols Na2O x (4 mols Na/2 mols Na2O) = about 0.06. Note that mols Na2O in the numeratorof the first term cancels with mols Na2O in the denominator of the second term which leaves mols Na which is what you want. In practice you ALWAYS know that the coefficient of the unit you are converting to goes on top and the coefficient of mols you have goes on the bottom.
4. Now convert mols of what you want to grams. grams = molsl x molar mass - about 0.06 x 23 = about ?
Copy this. It can come in handy for every chemistry course you take.
Explanation:
V=84.0 mL = 84.0 cm³
m=609.0 g
p=m/v
p=609.0/84.0=7.25 g/cm³
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:
E = m(ΔH)
where:
m = mass of substance
ΔH = latent heat of fusion or melting
According to data, the ΔH of mercury is approximately 11.6 kJ/kg.
E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J
The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
Answer:
E) A, B, and C
Explanation:
Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.
Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.