Answer:
Oxidation is the loss of electrons, that is, addition of electronegetive elements, example is addition of oxygen. Also, removal of electropositive elements, example is removal of hydrogen.
Explanation: a) In the presence of excess oxygen, propane burns in air, which gives the following chemical equation:
C3H8 + 5O2⇒ 3CO2 + 4H2O +Heat
b) When insufficient oxygen or too much oxygen is present for complete combustion, the following equation is given:
2C3H8 + 9O2 ⇒ 4CO2 + 2CO + 8H2O + Heat
c) At the anode( negative terminal): O∧2- ⇒ O + e
Oxygen accepts electron.
d) At cathode ( positive terminal): H∧+ + e∧- ⇒ H
Hydrogen donates electron
d) Nernst equation for reversal potential is given as follows:
E= RT/zF In{ion outside cell}/{ion inside cell}= 2.303 RT/zF In{ion outside cell}/{ion inside cell}
Percentage yield = actual yield/theoretical yield
So divide 21.0 g by 22.7 g and multiply it by 100 to find the percentage yield
Answer:
1 valence - Alkali Metals: Li Lithium, Na Sodium, K Potassium
2 valence - Alkaline Earth Metals: Be Beryllium, Mg Magnesium, Ca Calcium
3 valence - Non-metals: B Boron, Al Aluminium
4 valence - Non-metals: C Carbon, Si Silicon
5 valence - Non-metals: N Nitrogen, P Phosphorus
6 valence - Non-metals: O Oxygen, S Sulfur, Se Selenium
7 valence - Halogens: F Fluorine, Cl Chlorine, Br Bromine
8 valence - Noble Gases: He Helium, Ne Neon, Ar Argon
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
