The family on the periodic table that has a filled outer
energy level is VIIIA. The answer is letter D. They are also called the noble
gases or inert gases. They are virtually unreactive towards other elements or
compounds. They are found in trace amounts in the atmosphere. Their elemental form
at room temperature is colorless, odorless and monatomic gases. They also have
full octet of eight valence electrons in their highest orbitals so they have a
very little tendency to gain or lose electrons to form ions or share electrons with
other elements in covalent bonds.
Answer:
i think its true but I’m not sure
Explanation:
I know that they can. Be mixed
Answer:
metal : Mercury(Hg)
non metal : bromine (Br)
Explanation:
mercury is liquid at room temperature and pressure and the same as bromine
Answer:
d. is the hydrostatic pressure produced on the surface of a semi-permeable membrane by osmosis.
Explanation:
Osmosis -
It is the flow of the molecules of solvent from a region of higher concentration towards the region of lower concentration via a semipermeable membrane , is known as osmosis.
Osmotic pressure -
It refers to the minimum amount of pressure , which is required to be applied to the solution in order to avoid the flow of pure solvent via the semipermeable membrane , is referred to as osmotic pressure.
Or in simple terms ,
Osmotic pressure is the pressure applied to resists the process of osmosis.
Hence ,
From the given options in the question,
The correct option regarding osmotic pressure is d.
Answer:
See explanation below
Explanation:
In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.
At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.
Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.
In the picture below, you have the mechanism for all the 4 products.
Hope this helps
