The answer should be B: -3
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
Explanation:
c and d. I'm not sure which one is better tho
Answer: The statement, you've just prepared an aqueous solution is true.
Explanation:
When one or more number of substances are dissolved in a solvent like water then solution formed is called an aqueous solution.
For example, when a hot drink is made by dissolving a teaspoon of instant coffee and a teaspoon of sugar in a cup of hot water is an aqueous solution.
Here, both coffee and sugar are solute whereas hot water is the solvent.
Thus, we can conclude that the statement, you've just prepared an aqueous solution is true.
Answer:
1) Moles of CO 2 = Given mass / Molecular mass of CO 2 = 4.4 / 44 = 0.1 mole. 2) Molecules of CO2 in 0.1 moles of CO2 = 0.1 x 6.023 x 10 23 = 6.023 x 10 22 molecules 3) 44 gram (molecular wt of CO 2) contains 2 moles atom of oxygen therefore 4.4 gram of CO2 will contain = 2 /44 *4.4 = 0.2 moles atom of Oxygen.
