<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>
Answer:chicken butt I need some points
Explanation:blm ✊✊✊✊ white lives don't
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent Bond
Between 0.4 and 1.7 then it is Polar Covalent Bond
Greater than 1.7 then it is Ionic
For Br₂;
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar Covalent Bond)
For MgS;
E.N of Sulfur = 2.58
E.N of Magnesium = 1.31
________
E.N Difference 1.27 (Ionic Bond)
For SO₂;
E.N of Oxygen = 3.44
E.N of Sulfur = 2.58
________
E.N Difference 0.86 (Polar Covalent Bond)
For KF;
E.N of Fluorine = 3.98
E.N of Potassium = 0.82
________
E.N Difference 3.16 (Ionic Bond)
Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.
Answer:
a
Explanation:
it is A because h20 and c02 are molecules while what make them up are atoms
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
