Answer:
Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.
Explanation:
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
The answer is C............
You take the grams of CO₂ times Avogadro's number divided by the molar mass.
