<span>There are two kinds of forces, or attractions, that operate in a molecule so hydrogen would be the strongest attraction</span>
Answer:
4.22 g/mL
Explanation:
First we <u>calculate the mass of the solid</u> via <em>mass difference</em>:
- Mass of solid = 9.441 g - 1.005 g = 8.436 g
Then we <u>calculate the volume of the solid</u>, once again by difference:
- Volume of solid = 5.45 mL - 3.45 mL = 2.00 mL
Finally we <u>calculate the density in g/mL</u>:
- Density = 8.436 g / 2.00 mL = 4.22 g/mL
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
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For CaCl₂,
E.N of Chlorine = 3.16
E.N of Calcium = 1.00
________
E.N Difference 2.16 (Ionic Bond)
For C-H Bond in C₆H₁₂O₆,
E.N of Carbon = 2.55
E.N of Hydrogen = 2.20
________
E.N Difference 0.35 (Non-Polar Covalent Bond)
For C-O Bond in C₆H₁₂O₆,
E.N of Oxygen = 3.44
E.N of Hydrogen = 2.55
________
E.N Difference 0.89 (Polar Covalent Bond)
For O-H Bond in C₆H₁₂O₆,
E.N of Oxygen = 3.44
E.N of Hydrogen = 2.20
________
E.N Difference 1.24 (Polar Covalent Bond)
For MgO,
E.N of Oxygen = 3.44
E.N of Magnesium = 1.31
________
E.N Difference 2.13 (Ionic Bond)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For SiO₂,
E.N of Oxygen = 3.44
E.N of Silicon = 1.90
________
E.N Difference 1.54 (Polar Covalent Bond)
Result:
Compounds having Ionic Bonds are Na₂O, MgO and CaCl₂.
The value of ΔG° (gibbs free energy change) is -2703kj and -2657kj.
Given ,
combustion reaction of butane :
C4H10(g) + 13/2O2 (g) →4CO2(g) + 5H2O(g)
Method-1 :
We know ,
ΔGrxn = sum of ΔG (product ) - sum of ΔG (reactant )
= [4 × (-394.4) + 5×(-228.6) ] - [1×(-16.7)]
ΔGrxn = -2720.6kj
Method-2 :
We know ,
ΔG = ΔH -TΔS
T =298K
Thus , ΔHrxn = sum of ΔH (product ) - sum of ΔH (reactant )
= [4×(-393.5) + 5×(-241.8) ] - [ 1×(-126)]
ΔHrxn = -2657kj
ΔSrxn = sum of ΔS (product ) - sum of ΔS( reactant )
= [ 4×(213.7) + 5×188.7 ] - [ 13/2 ×205 + 1×310 ]
ΔSrxn = 155.8j/K
ΔSrxn = 0.1558kj/K
Thus , ΔG = ΔH - TΔS
= -2657 - ( 298 × 0.1558kj/K )
ΔG = -2720kj
Hence , the value of ΔG is -2703kj .
Learn more combustion reaction here :
brainly.com/question/13251946
#SPJ4