A cubic centimeter is equal in volume to one milliliter
Li + e- ===> Li+
it looses its valence electron to form a uni-positive ion
Answer:
First confirm the reaction is balanced:
C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).
a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.
b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).
There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.
MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g
c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.
The other questions use the same technique and will give you some much needed practice.
Explanation:
Answer:
B) exothermic.
Explanation:
Hello!
In this case, we need to keep in mind that exothermic reactions release heat, so they increase the temperature as the final energy is less than the initial energy; in contrast, endothermic reactions absorb heat, so they decrease the temperature as the final energy is greater than the initial energy.
In such a way, when a dissolution process shows off a negative enthalpy of dissolution, we infer it is an exothermic process due to the aforementioned; therefore, the answer is:
B) exothermic
.
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<u>Answer:</u> The entropy change of the process is 
<u>Explanation:</u>
To calculate the entropy change for different phase at same temperature, we use the equation:
where,
= Entropy change
n = moles of acetone = 6.3 moles
= enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = ![-94.7^oC=[273-94.7]=178.3K](https://tex.z-dn.net/?f=-94.7%5EoC%3D%5B273-94.7%5D%3D178.3K)
Putting values in above equation, we get:
Hence, the entropy change of the process is
