Answer:
2.445 g
Explanation:
Step 1: Given and required data
- Energy in the form of heat required to boil the water (Q): 5525 J
- Latent heat of vaporization of water (∆H°vap): 2260 J/g
Step 2: Calculate the mass of water
We will use the following expression.
Q = ∆H°vap × m
m = Q / ∆H°vap
m = 5525 J / (2260 J/g)
m = 2.445 g
MASS: generally a measure of an object's resistance to changing its state of motion when a force is applied. It is determined by the strength of its mutual gravitational attraction.
Molecular mass or molecular weight is the mass of a molecule.
Gram formula mass (a.k.a. molar mass) is defined as the atomic mass of one mole of an element, molecular compound or ionic compound. The answer must always be written with the unit g/mol (grams per mole).
(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
