Answer:
A. It would float with about 80% of the cube below the surface of the water and 20% above the surface.
Explanation:
The choice that best describes what happens to cube of the given density value is that it would float with about 80% of the cube would be below the surface of the water and 20% above the surface.
Density is the mass per unit volume of a substance. The more mass a body has relative to volume, the great it's density. In short, density is directly proportional to mass and inversely related to volume.
The density of water is 1g/mL
If the density of the cube were to be the same with that of water, the substance will just mix up with water .
Here the density is less than that of water.
The density is 0.2g/mL
Therefore, 20% will stay afloat and 80% will be below the surface of the water.
2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s
when the top of the pencil moves in one direction, the bottom end on the ground will move in another, but due to the frictional force it will move by a lesser amount, and the center of mass of the pencil will be accelerated in the horizontal direction due to the force of friction
Answer:
The answer to your question is: ΔHrxm = -23.9 kJ
Explanation:
Data:
2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ (1)
CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ (2)
Reaction:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
We invert (1) and change the sign of ΔH
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
We multiply (2) by 3
3( CO(g)+1/2O2(g)→CO2(g) ΔH = -282.7 kJ) (2)
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
We add (1) and (2)
Fe2O3(s) → 2Fe(s)+3/2O2(g) ΔH = 824.2 kJ
3CO(g)+3/2O2(g)→3CO2(g) ΔH = -848.1 kJ
Fe2O3(s) + 3CO(g)+3/2O2(g) → 2Fe(s)+3/2O2 + 3CO2(g)
Simplify
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) and ΔHrxm = -23.9 kJ
