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:
A
Explanation:
Hot air is less dense therefore it rises. Cold air is more closely packed therefore it is more dense. Less dense things rise while more dense things sink.
Answer:
Explanation:
Ionic bonds result from transfer of electrons, whereas covalent bonds are formed by sharing. 2. Ionic bonds are electrostatic in nature, resulting from that attraction of positive and negative ions that result from the electron transfer process; charge separation between covalently bonded atoms is less extreme.
Answer:
Specific heat of metal = 0.26 j/g.°C
Explanation:
Given data:
Mass of sample = 80.0 g
Initial temperature = 55.5 °C
Final temperature = 81.75 °C
Amount of heat absorbed = 540 j
Specific heat of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 81.75 °C - 55.5 °C
ΔT = 26.25 °C
540 j = 80 g × c × 26.25 °C
540 j = 2100 g.°C× c
540 j / 2100 g.°C = c
c = 0.26 j/g.°C
