methanol:
1 mole CH3 OH --> produces --> 1 mole CO2
1 mole CO2 has a molar mass of 44.01 gh/mole
your set up is:
(44.01 g CO2) / -726.5kJ = 0.06058g
your answer 0.06058 grams of CO2 produced per kJ released.
Answer:
Mark has a speed of 6 MPS (Miles Per Hour)
Explanation:
It took mark 2 hours to ride his bike to his grandma's house, which was 12 miles away. I divided 2 by 12 and got 6. Remember this: mph = miles away ÷ time.
The answer would be , a Chemical change!
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
