Bromide ions donates an electron in redox reactions.
<u>Explanation:</u>
- In these redox reactions, the halide ions like bromide donates a pair of electrons and acts as a reducing agents, but itself gets oxidized to bromine.
- In this process, the oxidation state of bromide ion is increased from -1 to 0 oxidation state, that is Br⁻ (-1) to Br₂ (0), thus reduces the compound and oxidizes by itself.
- Bromide ion is a strong reducing agent, thereby reduces sulfuric acid which changes to sulfur di oxide, but this doesn't happen in the case of chloride and fluoride ions as they are not having that much capacity like bromide and iodide ions.
Answer:Hope this helps!
Explanation:
You can use a flame test to help identify the composition of a sample. The test is used to identify metal ions (and certain other ions) based on the characteristic emission spectrum of the elements. The test is performed by dipping a wire or wooden splint into a sample solution or coating it with the powdered metal salt. The color of a gas flame is observed as the sample is heated. If a wooden splint is used, it's necessary to wave the sample through the flame to avoid setting the wood on fire. The color of the flame is compared against the flame colors known to be associated with the metals.
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
Answer:
pH = 10.75
Explanation:
To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]
Using the equation:
pH = 14 - pOH
We can find the pH of the solution.
The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M
pOH is
pOH = -log 5.6x10⁻⁴M
pOH = 3.25
pH = 14-pOH
<h3>pH = 10.75</h3>