Answer:
Here you can use the Clausis Clayperon equation: ln P1/P2=-Ea/R-(1/T1 - 1/T2)
where P1 is the pressure at standard condition: 760 mm Hg
P2 is the variable we need to solve
Ea is the activation energy, which in this case is delta H vaporisation: 56.9 kJ/mol
R is the gas constant 8.314 J/mol or 8.314 J/mol /1000 to convert to kJ
T1 is the normal boiling point 356.7 C, but converted to Kelvin: 629.85K
T2 is room temperature 25 C, but converted to Kelvin: 298.15 K
Once you plug everything in, you should get 4.29*10^-3 mmHg
Explanation:
Yes because I went through college and learned this
Answer:
-41. 47
Explanation:
m = q / Cp x T
m = Mass
q = Energy (or joules)
Cp = Heat Capacity
T = Change in Temperature
Water's heat capacity is always 4.18.
This is the formula you'll need for change in temperature:
Final - Initial
So, 33 - 78 = -45
m = 7800 / 4.18 x -45
= -41.47
Answer:
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
Explanation:
Step 1: Data given
2.0 M Al(NO3)3
Step 2:
Al(NO3)3 in water will dissociate as following:
Al(NO3)3 → Al^3+ + 3NO3^-
For 1 mol of Al(NO3)3 we will have 1 mol of Al^3+ and 3 moles of NO3^-
We know that the molarity of Al(NO3)3 = 2.0 M, this means 2.0 mol/ L
The mol ratio Al(NO3)3 and Al^3+ is 1:1 so the molarity of Al^3+ is<u> 2.0 M</u>
The mol ratio Al(NO3)3 and NO3^- is 1:3 so the molarity of NO3^- is<u> 6.0M</u>
The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)
