To increase the polarity of HBr, the bromine atom can be replaced with a hydrogen atom.
A polar molecule is one in which a dipole moment exists. There is a positive end and a negative end in a polar molecule. Conventionally, the direction of the dipole is from the positive end of the molecule towards the negative end of the molecule.
If we want to increase the polarity of the molecule then we must substitute the bromine atom with a more electronegative atom. In this case, chlorine is the best option.
Missing parts;
If you wanted to change the polarity of hydrogen bromide (HBr) by substitutingthe bromine by a different atom. Which atom would increase the polarity of the molecule?
A. chlorine (CI)
B. iodine (1)
C. sulfur (S)
D. hydrogen (H)
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Answer:
This question is incomplete as it lacks options, the options are:
A. chlorine (CI)
B. iodine (1)
C. sulfur (S)
D. hydrogen (H)
The answer is A. Chlorine
Explanation:
Polarity of a substance in chemistry is a function of electric charges in the atoms of the molecules involved. Polarity, however, can be increased or decreased in molecules depending on the charges of the atoms that form them.
Since polarity increases when an atom in the molecule has a high ability to pull electrons toward itself i.e. electronegativity, one atom that can be substituted for Bromine in the hydrogen bromide (HBr) molecule in order to increase its polarity is CHLORINE. This is because Chlorine (Cl) is more electronegative than Bromine atom, hence, will pull more electrons from hydrogen to make the HCl molecule more polar than HBr.
b)0.10 M NaCl vs. 0.10 M Al2(SO4)3 (assume complete dissociation of both salts).
c) 1.0 mole of NaCl added to water vs. 1.0 mole of pbCl2 added to water (hint: remember the solubility rules).
Answer:
Option 3.
Explanation:
Isomerism is a phenomenon where by two or more compounds have the same molecular formula but different structural patterns.
Geometric Isomerism is a type of Isomerism that occurs within a double bond i.e Geometric isomers have different arrangement within the double bond.
Considering the options given above,
The 1st option is exactly the same as the compound only, it is inverted.
The 2nd option is still the same as the compound, only it is laterally inverted.
The 3rd option satisfy geometric Isomerism as the arrangement differ from the compound in the double bond.
The 4th option is entirely a saturated compound in which geometric Isomerism is not possible.
Answer:
p orbitals only
Explanation:
Carbon has an atomic number of 6 so its electron configuration will be 1s² 2s² 2p². It has two orbitals as indicated with the 2 as its period number with the outer orbital have 4 valence electrons. So carbon is in the p-orbital, period 2 and in group 4.
Carbon's valence electrons reside in the 2s and 2p orbitals. These orbitals hybridize during bond formation to create equivalent sp3 hybrid orbitals, as evidenced in the methane molecule. Carbon's valence electrons are not placed in d orbitals.
Carbon (atomic number 6) has a total of six electrons. Two of these fill the 1s orbital. The next two fill the 2s orbital, and the final two are in the 2p subshell. According to Hund's rule, the most stable configuration for an atom is one with the maximum number of unpaired electrons. Therefore, carbon has two electrons in the 2s subshell and two unpaired electrons in two separate 2p orbitals. When discussing valence electrons, the electrons in the outermost shell are the ones considered, which for carbon are the electrons in the second shell namely 2s and 2p.
The geometry of the methane molecule (CH4) illustrates that in the bonding process, the s and p orbitals hybridize to allow the formation of four equivalent bonds with hydrogen atoms. Without hybridization, we would expect three bonds at right angles (from the p orbitals) and one at a different angle (from the s orbital). Nonetheless, through orbital hybridization, all four bonds in methane are identical, which is explained by the concept of sp3 hybridized orbitals.
Therefore, the valence electrons for carbon would be placed in the s orbital and p orbitals, not in the d orbitals, because carbon does not have electrons in the d subshell in its ground state. Additionally, the s and p orbitals are the only ones involved in bonding for carbon in most of its compounds, such as methane.
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Answer:
c. CH3CH2CH(CH3)CHO
Explanation:
Hello there!
In this case, according to the process for the one-step oxidation of a primary alcohol with a moderately strong oxidizing agent like pyridinium chlorochromate (PCC), whereby an aldehyde is produced, we infer that the corresponding product will be 2-methylbutanal, which matches with the choice c. CH3CH2CH(CH3)CHO according to the following reaction:
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Answer:
Nice pic there
Explanation:
No need