manganese The metal that is used as a sacrificial electrode to prevent the rusting of iron is manganese.
Answer : The final temperature of the mixture is, 
Explanation :
First we have to calculate the mass of ethanol and water.
and,
Now we have to calculate the final temperature of the mixture.
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of ethanol = 
= specific heat of water = 
= mass of ethanol = 35.5 g
= mass of water = 45.0 g
= final temperature of mixture = ?
= initial temperature of ethanol = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the final temperature of the mixture is,
I believe that the answer is ionic
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
This process is called the water cycle
