Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.
5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
Molar mass of LiBr (mm )= 86.845 g/mol
Molarity ( M ) = 4 M
Mass of solute ( m ) = 100 g
Volume ( V ) = in liters ?
V = m / mm * M
V = 100 / 86.845 * 4
V = 100 / 347.38
V = 0.2875 L
hope this helps!.
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
1 x 10^-4
Explanation:
Use the equation pH = -log[OH-}
Rearranging it [OH-] = 10^-pH
Plugging in we get [OH-] = 1 x 10^-4
We know that domestic dogs are descended from wolves. Exactly when and how this occurred is still being studied; many experts indicate that dogs were first domesticated about 12,000 years ago, but a few DNA studies indicate a much earlier date for the split between the wolf and the dog, perhaps as long ago as 130,000 years.
Regardless of the exact timing of the divergence between wolves and dogs, one thing is true: both species display quite a bit of genetic diversity, particularly in size and coloration.In order to breed a dog of a specific size, all one has to do is choose parents with the appropriate variations. For instance, if you want to create a small dog breed, you would just breed the smallest dogs you have, then you would choose their smallest offspring to breed the next generation, and so on until you have reached the desired size.