Oxidation-Reduction Reactions Suggested Reading Thus the oxidation number for oxygen in calcium oxide is -2. ... In effect, each calcium atom loses two electrons to form Ca2+ ions, and each O atom in O2 gains two electrons to form O2- ions. The net result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidation-reduction reaction.
+<u>O²</u><u>(</u><u>g</u><u>)</u><u>=</u><u>2</u><u>CaO</u><u>(</u><u>s</u><u>)</u>
Explanation:
we can conclude that in the reaction there is both reduction and oxidation.
In order to find the percent yield of a reaction, we need to use the theoretical yield and actual yield into a formula, which is the following:
%yield = (actual yield/theoretical yield)*100
Adding the values from the question in the formula:
%yield = (4.5/10.7)*100
%yield = 42
The percent yield for this reaction will be 42%
H2 <span>because the smaller the gas molecule, the faster the diffusion. (the lightest molecule will diffuse the quickest)</span>
base
bye hope you have a good day
Answer:
There are 0,2 moles of gas that ocuppy the container.
Explanation:
We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol. Firs we convert the unit of temperature in Celsius into Kelvin:
0°C= 273 K ------> 45,6 °C= 273 + 45, 6= 318, 6 K
PV= nRT ---> n= PV/RT
n= 1,48 atm x 3,45 L /0.082 l atm / K mol x 318,6 K
n= 0,195443479 mol
