Can you include the pome pls
Explanation:
According to Buoyance equation,
m = ![[m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]](https://tex.z-dn.net/?f=%5Bm%27%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd_%7Bw%7D%7D%7D%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd%7D%7D%5D)
where, m = true mass
m' = mass read from the balance = 17.320 g
= density of air = 0.0012 g/ml
= density of the balance = 7.5 g/ml
d = density of liquid octane = 0.7025 g/ml
Now, putting all the given values into the above formula and calculate the true mass as follows.
m =
= ![[17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]](https://tex.z-dn.net/?f=%5B17.320%20g%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B7.5%20g%2Fml%7D%7D%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B0.7025%7D%7D%5D)
= 
= 17.317 g
Thus, we can conclude that the true mass of octane is 17.317 g.
The answer is D.sucrose. Sucrose is a disaccharide .
Answer:
the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.
Explanation:
3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)
∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants
∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]
∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K
∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.
For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,
then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.
