<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.
The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
Answer:

Explanation:
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In this case, we are able to perform this unit conversion by cubing the relationship between inches and centimetres as shown below:

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Answer:
the chemical element of atomic number 16, a yellow combustible nonmetal.
Explanation:
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<span>1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2