Answer:
P4Se3
Explanation:
Applying the Law of Conservation of Mass:
(133 mg total) - (45.8 mg P) = 87.2 mg Se
Dividing by the molar mass,
Phosphorus:-(45.8 mg P) / (30.97376 g P/mol) = 1.4786 mmol P
Selenium:-(87.2 mg Se) / (78.96 g Se/mol) = 1.1044 mmol Se
Divide by the smaller number of millimoles:
(1.4786 mmol P) / 1.1044 mmol = 1.339
(1.1044 mmol Se) / 1.1044 mmol = 1.000
multiply by 3, to get a whole number interger. then round to the nearest whole numbers to find the empirical formula.
Therefore the empirical formula is P4Se3
Answer:
I would have to say Iron let me know if i am wrong
Have a great day and good luck
Answer: Hg
Explanation:
The symbol of Mercury is Hg drived from Hydra gayrum
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
Answer:
16.566g
Explanation:
The 0.223M barium bromide means there are 0.223 moles of barium bromide is 1000ml of barium bromide solution.
Now, we firstly need to know the number of moles in 250ml
And that is = (0.223 * 250)/1000 = 0.05575 mole
Now we need to know the amount in g of barium bromide to add.
From the relation : mass = number of moles * molar mass.
The molar mass of barium bromide is 297.14g/mol
mass = 297.14 * 0.05575 = 16.566g