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
The type of chemistry that allows the anthropologist to find out the nature of the substance in a mummy's wrap would be analytical chemistry. It is a branch of chemistry which deals with obtaining and processing information pertaining to the composition and the structure of the subject in question. It deals with apparatus and techniques to identify, separate and quantify a substance. It is the science of the determination of a substance and how much is present. By the analytical techniques, the anthropologist could hypothesize what is the substance of the wrap and its physical and chemical properties.
Explanation:
Number of moles(n)=Number of atoms(N)/Avogadro's constant.
Avogadro's constant=6.02×10²³
so we have
n=9.05×10²³/6.02×10²³
n=1.0503moles.
n=mass/molar mass
1.0503=mass/28
mass=1.0503×28
mass=29.4084g
Answer:
Boyles law
Explanation:
I think that is combined gas law.
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:
Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n
According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:
Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl: