The general trend in acidity among hydrohalic acids shows that HBr is stronger than HCl and weaker than HI. In summary, HBr is a prime example of a compound that meets both criteria: being a powerful acid and containing a halogen. Therefore, when looking for the compound that fits the description in the question, HBr is the correct answer.

HBr is a molecular compound because it is formed by the covalent bonding of non-metal atoms. When dissolved in water, it acts as an acid by dissociating to form hydrogen ions. The correct classification for HBr is a molecular compound that is an acid. Thus, the correct answer is option 3) Molecular compound, acid. Explanation:

Determine the name for aqueous HBr. a. bromous acid b. hydrogen bromate c. bromic acid d. hydrobromic acid e. hydrobromous acid; Classify the compound given as a strong acid, weak acid, strong base, or weak base. H_2SO_4. For the compound NaCl, identify which of the following applies. a. ionic b. molecular c. acid d. base

spell out the full name of the compound: HBr. There are 2 steps to solve this one. Solution. Step 1. HBr ...

2023年10月31日 · Understanding HBr in Compound and Acidic Forms: When discussing HBr in its compound form, it is referred to as hydrogen bromide. Hydrogen bromide is a diatomic molecule consisting of one hydrogen atom and one bromine atom, represented chemically as HBr. In its acidic form, HBr is known as hydrobromic acid.

The proper name for HBr (aq) is hydrobromic acid. In chemistry, when hydrogen combines with a halogen (a group 7 element), the resulting compound is named as hydro- + [halogen name] + ic acid. In this case, the halogen is bromine, so the compound is named hydrobromic acid.

(a) Name the compound: Au_2S_2O_3. (b) Identify it as an ionic compound, acid, or molecular compound. Give the name and formula of the binary compound formed by Na and H. Is the compound with the formula Mg(NO3)2 ionic, molecular, or an acid? Write its name. Is the compound with the formula CBr4 ionic, molecular, or an acid? Write its name.

Hydroboration-oxidation of compound A produces a pair of enantiomers, compounds B and C. When treated with HBr, compound A is converted into compound D, which is a tertiary alkyl bromide. When treated with O3/DMS, compound A is converted into compounds E and F. Compound E has three carbon atoms, while compound F has only two carbon atoms.

Question: Compound A reacts with HBr/ROOR to give compound B. Compound C reacts with bromine and light to produce the same compound (compound B). What are likely structures for compounds A, B, and C? Br -Br = III IV O Compound Ais II, Compound B is IV, and Compound C is III O Compound A is II, Compound B is I, and Compound C is IV O Compound A ...

The compound that reacts the fastest with HBr is 1, while the compound that reacts the slowest with HBr is 5. Here come energy that rest hefiates with Br is , while ooo or on of Your solution’s ready to go!