if I'm not mistaking it would be the second option
Answer:
C₆H₆O₃
Explanation:
Calculation sequence:
% => grams => moles => reduce => empirical Ratio
Molecular multiple = Molecular Mass / Empirical Mass
C: => 57.1% => 57.1 g => 57.1/12 = 4.7583
H: => 4.8% => 4.8 g => 4.8/1 = 4.8000
O: => <u>38.1% => 38.1 g </u>=> 38.1/16 = 2.3813
TTL => 100% 100 g
Reduced Mole values =>
C : H : O => 4.7583/2.3813 : 4.8000/2.3813 : 2.3813/2.3813 => 2 : 2 : 1
∴ empirical formula => C₂H₂O
empirical formula weight => 2C + 2H + 1O = [2(12) + 2(1) + 1(16)] amu = 42 amu
molecular formula weight (given in problem) = 126 g/mole
The molecular formula is a whole number multiple of the empirical formula.
molecular multiple = 126 amu / 42 amu = 3
∴ molecular formula => (C₂H₂O)₃ => C₆H₆O₃
Answer:
that squirrel would be dead if it was my mom
Explanation:
The molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol, Option D is the correct answer.
<h3>What is enthalpy of Combustion ?</h3>
The energy released when a fuel is oxidized by an oxidizing agent is called enthalpy of Combustion.
It is given that
a 1.0 g sample of magnesium is burned to form MgO. in doing so, 25.5 kj of energy are released.
Molecular weight of Magnesium = 24.35g
24.35 g makes 1 mole of Mg
1g = 1/24.35
For 0.04 moles 25.5 kJ is released
for 1 mole 25.5 *1/.04
= 620 kj/mol
Therefore the molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol.
To know more about enthalpy of combustion
brainly.com/question/14754029
#SPJ1
