Answer:
Explanation:
(a) Answer: Intermolecular forces
The reason for this answer is because the substance (paraffin wax) only changed it's state from solid to liquid and didn't undergo a breakage in it's covalent bond within it's carbon chain which would have produced another substance.
(b) Solid substances are generally more dense than there corresponding liquid substances because the more compact particles are (which occurs in solids), the more dense they become. They are thus more dense than liquids because liquids have there particles loosely packed and well spaced making them less dense than there corresponding solids. Hence, the solid paraffin wax was going to become less dense because it's particles moved from being tightly packed (as solids) to being loosely packed (as liquids). Density refers to mass per volume but can also be described as the level of compactness of a substance. Thus, since liquid is not as compact as solid, it can be said to be less dense than solids.
Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature =
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 = 
C = 0.0285
This also means that,
= 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
= 
Now, let us assume that mass of
= x grams
And, mass of
= (1.00 - x)
So, moles of
=
Now, moles of
=
=
= x = 0.346
Therefore, we can conclude that amount of
present is 0.346 g and amount of
present is (1 - 0.346) g = 0.654 g.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>