The heat will flow from copper to aluminum because Cu is at higher temperature. The heat liberated is -7.60kJ
When two metals at different temperatures are kept in contact, heat flows from hotter metal to colder metal until thermal equilibrium is reached.
Here Copper is at a temperature of 60 degree Celsius and aluminum is at 40 degree Celsius. Thus, heat will flow from Cu to Al.
In order to calculate the amount of heat liberated following calculations are required.
m1=262 g
T1=87 oC
Cp=0.385 J/g oC
T2=11.8 oC
The heat liberated can be expressed as follows:
Q=mCp(T2-T1)
Q=262 g*0.385 J/goC(11.8-87)oC
Q=-7585 J
=-7.60kJ
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FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :

To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
Answer:
Molecules move from areas of high concentration to areas of low concentration.
1 km is equal to 100,000 cm

and you need to divide it to 1 km so that you can get rid of the km.