Answer:
EMPIRICAL FORMULA:
C₂H₃O₂
Explanation:
Mass of compound = 4.647 g
Mass of CO₂ = 8.635 g
Mass of H₂O = 1.767 g
Empirical formula = ?
Solution:
Percentage of C = 8.635/ 4.647 × 12/44 ×100
Percentage of C = 1.86× 12/44 ×100 = 50.7
Percentage of H = 1.767/4.647 × 2/ 18 × 100
Percentage of H = 0.38 × 2/ 18 × 100
Percentage of H = 4.18
Percentage of O = 100 - (50.7+4.18)
Percentage of O = 100 - 54.88
Percentage of O = 45.12
Number of grams atom:
Number of grams atom of C = 50.7/ 12 = 4.23
Number of grams atom of H = 4.18 /1 = 4.18
Number of grams atom of O = 45.12/ 16 = 2.82
Atomic ratio:
C : H : O
4.23/2.82 : 4.18/2.82 : 2.82/2.82
1 : 1.5 : 1
C : H : O (1 : 1.5 : 1)
C : H : O 2(1 : 1.5 : 1)
C : H : O (2 : 3 : 2)
EMPIRICAL FORMULA:
C₂H₃O₂
Answer: B) -4819 kJ
Explanation:
The balanced reaction for combustion of octane is:
![\Delta H^0_{rxn}=-11018kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5E0_%7Brxn%7D%3D-11018kJ)
To calculate the number of moles, we use the equation:
given mass of octane = 100.0 g
Molar mass of octane = 114.33 g/mol
Putting in the values we get:
![\text{Number of moles}=\frac{100.0g}{114.33g/mol}=0.8747moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B100.0g%7D%7B114.33g%2Fmol%7D%3D0.8747moles)
According to stoichiometry:
2 moles of octane give heat = -11018 kJ
Thus 0.8747 moles of octane give =
Thus -4819 kJ of heat is released by 100.0 g of octane assuming complete combustion.
Answer:
Principal Quantum Number (n) = energy level (1s2)
Angular Momentum Quantum Number (<em>l</em>) = sub-shell (1s2)
Magnetic Quantum Number (m<em>i</em>) = specific orbital electron occupies
range of (n) = 1-7
range of (<em>I</em>) = s, p, d , f
Explanation:
Sodium mn2 and po4 makes sodium
Symbol : O
Element Family : Oxygen Group or Chalcogens: Group 16 (VIA) - 6 valence electrons
Atomic Number : 8
Atomic Mass : 15.999 u