Answer:
The electron configuration for this atom is Calcium, which has 2 valence electrons.
Explanation:
Following the periodic table and with the electron configuration, you will end up with calcium, which has 2 valence electrons. (Always follow the electron configuration from left to right! It begins at hydrogen, then to helium... and so on.)
1s2 -> He....
2s2 -> Be....
2p6 -> Ne...
3s2 -> Mg...
3p6 -> Ar...
4s2 -> Ca.
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Answer:
Purpose: To become familiar with the techniques for separation of amixture of solids.
Explanation:
a mixture of pure substances. If you have a mixture of tennis ballsand marbles (not pure substances by the way), it would be easy toseparate the mixture. However, it is more difficult to separate asand (also not a pure substance) and salt mixture. Even with verygood tweezers and a magnifying glass, it would be extremelytedious. You could take advantage of the fact that salt dissolvesin water and sand does not. To separate iron powder from an ironand sand mixture you can take advantage of the magnetic propertiesof iron and separate the mixture.
To summarize a complete procedure for separating a mixture ofseveral substances, it is best to prepare a flow chart. A flowchartis a schematic representation of an algorithm or a stepwiseprocess, showing the steps as boxes of various kinds, and theirorder by connecting these with arrows. Flowcharts are used indesigning or documenting a process.
An atom of carbon has 4 electrons in its outermost shell, which means that
<span>its ionic charge is 4+ or 4-
</span>Si is in same group as carbon so its also 4+ or 4-
Germanium is 4+.
Sn is also 2+ or 4+
Pb is usually +2