Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:


We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

Now we can <u>put the values into the equation</u>:

Now we can <u>solve for the initial temperature of gold</u>, so:

ºC
I hope it helps!
The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S
Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol
If we take a basis of 1 mol, the answer is
D. -170 kJ
The group is might be labeled as VIIB or VIIA.
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>