Dipole-Induced-Dipole is the strongest type of intermolecular force between solute and solvent.
<h3>What is Intermolecular Force ?</h3>
Intermolecular force is also called secondary force is the force of attraction between molecules. It acts between ions and atoms.
<h3>What is Dipole-Induced-Dipole attraction ?</h3>
A dipole-induced-dipole attraction is a weak attraction it occurs when the partial charge form with in the molecule due to uneven distribution of charge in a molecule.
CCl₄ is non polar in nature and CH₃OH is polar in nature so dipole-induced-dipole attraction is present.
Thus from the above conclusion we can say that Dipole-Induced-Dipole is the strongest type of intermolecular force between solute and solvent.
Learn more about the Dipole-Induced-Dipole here: brainly.com/question/22973877
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Answer: Sodium Chloride because its still a liquid at the 773 temperature
Explanation: hope this helps :)
Answer:
The symbol (l) stands for liquid phase.
Explanation:
Answer:
The question is incorrect and incomplete. Here's the correct question:
It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy when burned. To illustrate this difficulty,a) calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 23.5 °C to 100 °C , it boils, and the resulting steam is raised to 315 °C. b)Discuss additional complications caused by the fact that crude oil has less density than water.
Explanation:
Q= mc ΔT
Q= heat energy
m is mass
ΔT is change in temperature and c is specific heat capacity
calculating heat for latent heat of vaporisation
Q= ml where l is latent heat of vaporisation
a) Total heat energy used= heat required to raise temperature from 23.5 °C to 100 °C, heat required to boil water and heat required to further raise temperature from 100 °C to 315°C
Q = mc ΔT₁ + mL + mc ΔT₂
Q = m(c ΔT₁ + L + c ΔT₂)
m= Q÷(c ΔT₁ + L + c ΔT₂)
Q= 2.8 X 10⁷ J
c=4186J/kg°C
L=2256 x 10³J/kg
ΔT₁=76.5°C(100°C-23.5°C)
ΔT₂= 215°C(315°C-100°C)
(c ΔT₁ + L + c ΔT₂)= 4186J/kg°C *76.5°C + 2256 x 10³J/kg + 4186J/kg°C*215°C =3476219J/Kg
m= 2.8 x 10⁷J ÷3476219J/Kg
m =80.54 Kg
volume = mass÷ density
=80.54kg ÷ 10³kg/m³( density of water)
=0.0854m³
0.001m³ = 1 lL0.08054m³= 0.08054m³ /0.001m³= 80.54L
VOLUME is 80.54litres
b) since the density of crude is less than the density of water,and 80L of additional water is added, it'll make the crude to float on water thus inhibiting the extinguishing process
Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
