In chemistry, a valence electron is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
The compounds that are present in both waxes and sphingomylin are FATTY ACID AND LONG CHAIN ALCOHOL.
Waxes are diverse type of organic substances which are hydrophobic in nature and they are found in both plant and animals. Sphingomyelin is a type of sphingolipid, which are found in animal cell membranes. It is especially abundant in the membranous myelin sheath which surround some nerve cells.<span />
A, G, E, H, F are the answers you desire. Those are the proper locations for the five statements about the carbon cycle.
Answer:
See explanation and images attached
Explanation:
There are two compounds that could have molecular formula of C2H6O, they are ethanol and methoxy methane (trivial name: methyl ether). The structures of the two compounds are shown in the image attached as obtained from quora.
The boiling point of methoxymethane is -24.8°C while that of ethanol is about 78.5°C. The reason for the high boiling point of ethanol is the presence of intermolecular hydrogen bonding leading to a stronger intermolecular interaction compared to methoxymethane molecules held together only by very weak dispersion forces.
Answer:
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺
Explanation:
Step 1: Write the unbalanced reaction
BrO₃⁻ + Sb³⁺ ⟶ Br⁻ + Sb⁵⁺
Step 2: Identify both half-reactions
Reduction: BrO₃⁻ ⟶ Br⁻
Oxidation: Sb³⁺ ⟶ Sb⁵⁺
Step 3: Perform the mass balance, adding H⁺ and H₂O where appropriate
6 H⁺ + BrO₃⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺
Step 4: Perform the charge balance, adding electrons where appropriate
6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻
Step 5: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost is the same
1 × (6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O)
3 × (Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻)
Step 6: Add both half-reactions and cancel what is repeated in both sides
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺