Answer:
The correct answer is V1/T1=V2/T2.
Explanation:
Just took the test
Answer:
w = -531 kJ
1. Work was done by the system.
Explanation:
Step 1: Given data
- Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
- Change in the internal energy of the system (ΔU°): 156 kJ
Step 2: Calculate the work done (w)
We will use the following expression.
ΔU° = q + w
w = ΔU° - q
w = 156 kJ - 687 kJ
w = -531 kJ
By convention, when w < 0, work is done by the system on the surroundings.
Answer:
The answer you are looking for is A
P=nRTV
hope this help<span />
Answer:

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+