It looks all correct to me, great job!
Copper heat capacity would be <span>0.385J/C*gram which means it needs 0.385 Joule of energy to increase 1 gram of copper temperature by 1 Celcius. The calculation would be:
energy= heat capacity *mass * temperature difference
energy= </span>0.385J/C*gram * 6g * (90-20)
<span>energy= 161.7J
</span>
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 - 
Solving
-
=75.5 kJ/mol - 2*90.25 kJ/mol
-
=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)