The mantle layer is more dense closest to the core.
<u>Given:</u>
Mass of ice = mass of water = 5.50 kg = 5500 g
Temperature of ice = -20 C
Temperature of water = 75 C
<u>To determine:</u>
Mass of propane required
<u>Explanation:</u>
Heat required to change from ice to water under the specified conditions is:-
q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)
= m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)
= 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ
The enthalpy change for the combustion of propane is -2220 kJ/mol
Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane
Molar mass of propane = 44 g/mol
Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g
Ans: 75.15 grams of propane must be combusted.
Answer:
2.60 moles of A remaining.
Explanation:
According to Le Chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.
In this question, we were told that the volume doubles, that implies that we would have to double the molarity of B/ C (since B=C.)
However, it is obvious and clear from the given equation of the reaction that A is solid in it's activity = 1. Hence, it is then ignored.
So doubling B would be 1.30 M × 2 = 2.60 M
i.e 2.60 M moles of A was consumed.
Now; the number of moles of A remaining is 5.20 - 2.60 = 2.60 moles of A remaining.
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
Answer:
S₁₂
Explanation:
The freezing point depression (ΔTf) is a colligative property that can be calculated using the following expression.
ΔTf = Kf × m
where,
Kf: freezing point depression
m: molality
ΔTf = Kf × m
m = ΔTf / Kf
m = 0.156 °C / (29.8 °C/m)
m = 5.23 × 10⁻³ m
The molality is:
m = moles of solute / kilograms of solvent
moles of solute = m × kilograms of solvent
moles of solute = 5.23 × 10⁻³ mol/kg × 0.5000 kg
moles of solute = 2.62 × 10⁻³ mol
1.00 g corresponds to 2.62 × 10⁻³ moles. The molar mass of Sₙ is:
1.00 g/2.62 × 10⁻³ mol = 382 g/mol
We can calculate n.
n = molar mass of Sₙ / molar mass of S
n = (382 g/mol) / (32.0 g/mol)
n = 11.9 ≈ 12
The molar formula is S₁₂.
