<span>The "second" is the SI base unit of time.</span>
Answer : The mass of needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of .
Now we have to calculate the moles of .
The balanced chemical reaction will be,
produced from 1 mole of
So, 0.005 mole of produced from 0.005 mole of
Now we have to calculate the mass of
Therefore, the mass of needed are, 1.515 grams.
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:
2. All the naturally occurring isotopes of Mg.
Explanation:
You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.
1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.
4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.
Answer : The molar mass of the solute will be
87.90 g/mol.Explanation : We know the formula for elevation in boiling point, which is
Δt = i
m
given that, Δt = 0.357,
= 5.02 and mass of
= 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X
.
Now, 0.250/ 2.844 X
=
87.90 g/mol. which is the weight of unknown component.