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.
The amount of Al2O3 in moles= 1.11 moles while in grams = 113.22 grams
<em><u>calculation</u></em>
2 Al + Fe2O3 → 2Fe + Al2O3
step 1: find the moles of Al by use of <u><em>moles= mass/molar mass </em></u>formula
= 60.0/27= 2.22 moles
Step 2: use the mole ratio to determine the moles of Al2O3.
The mole ratio of Al : Al2O3 is 2: 1 therefore the moles of Al2O3= 2.22/2=1.11 moles
Step 3: finds the mass of Al2O3 by us of <u><em>mass= moles x molar mass</em></u><em> </em>formula.
The molar mass of Al2O3 = (2x27) +( 16 x3) = 102 g/mol
mass is therefore= 102 g/mol x 1.11= 113.22 grams
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Answer:
Following laboratory safety guidelines minimizes the chance of lab accidents.
Explanation: