Answer:
8740 joules are required to convert 20 grams of ice to liquid water.
Explanation:
The amount of heat required (
), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:
![Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]](https://tex.z-dn.net/?f=Q%20%3D%20m%5Ccdot%20%5Bc%5Ccdot%20%28T_%7Bf%7D-T_%7Bo%7D%29%2BL_%7Bf%7D%5D)
(1)
Where:
- Mass, measured in grams.
- Specific heat of ice, measured in joules per gram-degree Celsius.
, 
- Temperature, measured in degrees Celsius.
- Latent heat of fussion, measured in joules per gram.
If we know that 
, 
, 
, 
and 
, then the amount of heat is:
![Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}](https://tex.z-dn.net/?f=Q%20%3D%20%2820%5C%2Cg%29%5Ccdot%20%5Cleft%5C%7B%5Cleft%282.06%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%5B0%5C%2C%5E%7B%5Ccirc%7DC-%28-50%5C%2C%5E%7B%5Ccirc%7DC%29%5D%2B334%5C%2C%5Cfrac%7BJ%7D%7Bg%7D%20%5Cright%5C%7D)
8740 joules are required to convert 20 grams of ice to liquid water.
Pollution and rising temperatures
0.500 moles is roughly .5*6.022*10^23=3.011*10^23 atoms. This is independent of STP.
Answer:
We need 41.2 L of propane
Explanation:
Step 1: Data given
volume of H2O = 165 L
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 3: Calculate moles of H2O
1 mol = 22.4 L
165 L = 7.37 moles
Step 4: Calculate moles of propane
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane
Step 5: Calculate volume of propane
1 mol = 22.4 L
1.84 moles = 41.2 L
We need 41.2 L of propane
