<u>Answer:</u>
CHCl3 has dipole-dipole interactions.
<u>Explanation:</u>
Trichloromethane has a electric dipole moment permanently pointing along the line parallel with the Hydrogen carbon axis.Dipole-dipole interactions are said to be intermolecular force of attractions that is formed from two permanent dipoles interacting.
These type of interactions are occurring when one of the partially charged formed molecule are being attracted to an opposite partially charged molecule nearby . The molecules align in a state that the positive end of one of the molecule gets interacting with the negative end of the another molecule.
It's 3. There are two Cl in the reactants and 3 in the products, and they need to be balanced/equal. Only way to do that is add three on the left side, which will give you 6 Cl (3×2Cl) and then 2 on the other side, which will also give you 6 Cl (2×3Cl)
The graph that shows that gases become increasingly soluble in a liquid as the temperature of the liquid decreases is a graph that shows these characteristics:
1) vertical-axis (traditional y-axis): solubility
2) horizontal-axis (traditional x-axis): temperature
3) graph: a curve with negative slope, i.e. the as the temperature (horizontal-axis) goes further to the right (increasing), the solubility (vertical-axis) goes downward.
An example of the application of this property is the CO2 in the sodas. CO2 incorporated at low temperature.
Answer:
An aluminium drink can.
24 carat gold.
The helium gas in a balloon that floats upwards.
Cast iron garden railings.
Lead sheeting used by builders on the roofs of houses.
Answer:
1.54 moles of pure ethanol are present in 750mL of wine
Explanation:
The sort of concentration 12% by volume, means that in 100 mL of solution (wine) 12 mL are pure ethanol
We use the density of pure ethanol to determine the mass:
Ethanol density = Ethanol mass / Ethanol volume
Ethanol density . Ethanol volume = Ethanol mass
0.789 g/mL . 12 mL = 9.47 g of ethanol
Now we can determine the moles: 9.47 g . 1 mol / 46.06 g = 0.205 moles
Now we can make a rule of three:
In 100 mL of solution (wine) we have 0.205 moles of pure ethanol
Then, in 750 mL of solution we may have (750 mL . 0.205) / 100 = 1.54 moles