Is four electrons the answer you’re looking for?
<u>Answer</u>
So this is the reaction that happens.
<span>C4H10 + O2 = CO2 + H2O </span>
<span>Balanced, it is </span>
<span>2C4H10 + 8O2 = 8CO2 + 10H2O </span>
<span>Given 1 kg or 1000 g of butane, use stoichiometry aka factor labeling aka conversions and mole ratios to get to grams of oxygen. </span>
<span>I'll do an example. Let's form water. Hydrogen is diatomic too. </span>
<span>2H2 + O2 = 2H2O </span>
<span>Given 1000 g of Hydrogen, I need to know how many grams of oxygen to use. To convert grams to moles,
I know that 1 mol of H2 equals 2.02 g. Then, for every mole of O2, there are 2 moles of H2. Then converting moles of O2 to grams, I know that one mole of it equals 32 grams. </span>
<span>[1000 g H2] x [1 mol H2/2.02 g H2] x [1 mol O2/2 mol H2] x [32 g O2/1 mol O2] </span>
<span>My answer would be 7.9 kg </span>
As per the rule, oxidation number of alkaline earth metals in their compounds is +2. Oxidation number of oxygen in it's compounds is -2(except peroxides) and the sum of oxidation numbers of all the elements of a neutral compound is zero.
Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
