Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
<span>Answer is: the symbol is Cl.
[Ne ] 3s</span>² 3p⁶ is electric configuration of noble gas argon, neon (Ne) has10 electrons plus 6 electrons in 3s and 3p orbitals. Neutral atom of m<span>onatomic ion that has a charge of 1– has one electron less than argon, so that atom (chlorine) has 17 electrons. Charge of 1- means one electron more for ion: 17 + 1 = 18.
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Answer:
think I did this before and its V
Answer:
23.0 s⁻¹ is rate constant
Explanation:
Using the Arrhenius equation:
k = A * e^(-Ea/RT)
Where k is rate constant
A is frequency factor (1.5x10¹¹s⁻¹)
Ea is activation energy = 55800J/mol
R is gas constant (8.314J/molK)
And T is absolute temperature (24°C + 273 = 297K)
Replacing:
k = 1.5x10¹¹s⁻¹ * e^(-55800J/mol/8.314J/molK*297K)
k = 1.5x10¹¹s⁻¹ * 1.53x10⁻¹⁰
k = 23.0 s⁻¹ is rate constant i hope this helpsss
Explanation:
