Answer:
1367.7 g of ethylene glycol was added to the solution
Explanation:
In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')
525.8 mmHg - 451 mmHg = 451 mmHg . Xm
74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)
Xm = Mole fraction of solute / Moles of solute + Moles of solvent
We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol
(Notice we converted the 2kg of water to g)
0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent
0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute
18.4 moles = Moles of solute - 0.166 moles of solute
18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles
Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g
Answer: 193 mg of theobromine are present in the sample.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.
To calculate the moles, we use the equation:
1 mole of theobromine 
weigh = 180 g
of theobromine weigh = 
(1g=1000mg)
193 mg of theobromine are present in the sample.
Answer:
the answer will be Oa. 2.26
Answer: A. -396 kJ
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Reversing the reaction, changes the sign of 
On multiplying the reaction by 2, enthalpy gets multiplied by 2:
Thus the enthalpy change for the reaction is -396 kJ.
