27*9=243 if one mole is equal to 27 grams times that by 9
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O
Answer:
A) 0.20 cm³
B) 49.7 m²
C) 99.99%
D) 17.7 mg
Explanation:
A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):
d =m/v
If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:
0.20 = 0.04/v
v = 0.04/0.20
v = 0.20 cm³
B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):
S = a/m
If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:
1242 = a/0.04
a = 49.7 m²
C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:
%removed = [(7.748 - 0.001)/7.748] *00%
%removed = 99.99%
D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:
m = 10 + 7.747
m = 17.747 mg
m = 17.7 mg