718.65 degrees is the initial temperature of the zinc metal sample.
Explanation:
Data given:
mass of zinc sample = 2.50 grams
mass of water = 65 grams
initial temperature of water = 20 degrees
final temperature of water = 22.5 degrees
ΔT = change in temperature of water is 2.50 degrees
specific heat capacity of zinc cp= 0.390 J/g°C
initial temperature of zinc sample = ?
cp of water = 4.186 J/g°C
heat absorbed = heat released (no heat loss)
formula used is
q = mcΔT
q water = 65 x 4.286 x 2.5
q water = 696.15 J
q zinc = 2.50 x 0.390 x (22.50- Ti)
equating the two equations
696.15 = - 22.50+ Ti
Ti = 718.65 degrees is the initial temperature of zinc.
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
Answer: V= 3.13 L
Explanation: solution attached:
Use combine gas law equation:
P1 V1 / T1 = P2 V2/ T2
Derive to find V2
V2 = P1 V1 T2 / T1 P2
Convert temperatures in K
T1= 13.0°C + 273 = 286 K
T2= 22.5°C + 273 = 295.5 K
Substitute the values.
Convert 57.6 L to dm3 and divide it by 24
Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate 
in order to prepare a 0.452 m solution
Explanation:
Molality : It is defined as the number of moles of solute present per kg of solvent
Formula used :
where,
n= moles of solute
Moles of 
= weight of the solvent in g = ?
Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate in order to prepare a 0.452 m solution
