All of the lower case letters are recessive
Answer:
-411 kj
Explanation:
We solve by using this formula
∆U = ∆Q + ∆W
This formula is the first law of thermodynamics
Change in internal energy U = +241
Heat gained by system Q = 652
Putting the value into the equation
+241 = 652 + W
Workdone = 241 - 652
Workdone = -411 kj
Since work done is negative it means that work was done by the system
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
Answer:
a. The pressure in the flask open to the atmosphere during the vaporization of the unknown liquid is equal to the prevailing atmospheric pressure equivalent to 0.957734 atm
Explanation:
The mass of the empty flask and stopper, m = 53.256 g
The volume of the unknown liquid she adds = 5 mL
The temperature of the water in which she heats up the flask = 98.8 °C = 371.95 K
The mass of the flask and the condensed vapor = 53.780 g
The volume of the flask, V = 231.1 mL
The atmospheric pressure, P = 728 mm Hg
a. We note that the student stoppers the flask after all the liquid has evaporated. Therefore, given that the flask was open to the atmospheric pressure as the liquid evaporates, the pressure of the vapor in the flask is equal to the prevailing atmospheric pressure, or 728 mmHg
Using a calculator, 728 mm Hg is equivalent to 0.957734 atm.