Depends on what the base is. You would reference the base dissociation chart for that value.
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
The finagling in the hole
Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm
