D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
Answer:
It is classified as a pure substance
Explanation:
Oxygen is a gas present in the atmosphere in the pure elemental form.
Answer:
Explanation:
(a) Balanced equation
2Zn + O₂ ⟶ 2ZnO
(b). Calculation
You want to convert moles of ZnO to moles of Zn
The molar ratio is 2 mol Zn:2 mol ZnO
First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4
