The three mass value measure are precise mass
<u>explanation</u>
precise mass is term use to describe data from experiment that have been repeated several times. An experiment that yield tightly grouped set of data it has a high precision. 8.93 , 8.94 and 8.92 are precise mass since they have repeated severally
I believe the answer is C
The full question is shown in the image attached
Answer:
See explanation
Explanation:
In naming an alkane, the first thing we do is to obtain the parent chain by counting the number of carbon atoms in the chain.
When we obtain that, then we identify the substituents and number them in such a way that they have the lowest numbers. The compounds shown have the following names according to the order in which the structures appear in the image attached;
1. 2-methyl propane
2. 2,4-dimethyl heptane
3. 2,2,3,3-tetramethyl butane
4. 5-ethyl-2,4-dimethyl octane
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
