Answer:
Explanation:
Hello!
In this case, according to the following chemical reaction:
It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:
Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:
Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:
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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:
2.25 g
Explanation:
The mass of the solid X must be the total mass (beaker + solid X) less than the mass of the beaker. Then:
mass of the solid X = 34.40 - 32.15
mass of the solid X = 2.25 g
The difference of 0.25 g must occur for several problems: an incorrect weight in the balance, the configuration of the balance, the solid can be hydrophilic and absorbs water, and others.
Answer:
No
Explanation:
The mass fraction is defined as:
where:
- wi: mass fraction of the substance i
- mi: mass of the substance i
- mt: total mass of the system
<u><em>The mass fraction of two substances (A and B), will be the same, ONLY if the mass of the substance A (mA) is the same as the mass of the substance B (mB).</em></u>
An equimolar mixutre of O2 and N2 has the same amount of moles of oxygen and nitrogen, just to give an example let's say that the system has 1 mole of O2 and 1 mole of N2. Then using the molecuar weigth of each of them we can calculate the mass:
mA= 1 mole of O2 * 16 g/1mol = 16 g
mB=1 mole of N2 *28 g/1mol=28 g
As mA≠mB then the mass fractions are not equal, so the answear is NO.
Answer:
The IUPAC structure only shows bond pairs and lone pairs. In the flouromethane structure above, there is only one bond pair and three lone pairs of electrons. Therefore there is one electron remaining, but since it doesn't not make up a pair, it is ignored in the structure but theoretically it is present.
