Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>
Answer 19.9g. I’ve took the test last week at my uncle randy’s house
.energy mechanical needs it Answer:
Yes.
Explanation:
Explanation:
As it is known that there are two types of properties. These are extensive and intensive.
Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.
Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.
On the basis of these properties water and ethanol are distinguished as follows.
- Density of water is 997 kg/
whereas density of ethanol is 789 kg/. Both these liquids can be separated by intensive properties. - Melting point of water is zero degree celsius whereas melting point of ethanol is -114.1 degree celsius.
- Specific heat capacity of water is 4.184
whereas specific heat capacity of ethanol is 2.46 . - Mass of the given liquids cannot be differentiated because they will keep on changing depending on the quantity required. As mass is an extensive property, therefore, it is difficult to differentiate between the two liquids.
Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.
