Mass of water vapor produced : 0.90 g
<h3>Further explanation</h3>
Given
0.10 g of hydrogen reacted with 0.80 g of oxygen
Required
mass of water vapor produced
Solution
Reaction
2H₂ + O ⇒ 2H₂O
If we refer to the law of conservation of mass which states that the mass before and after the reaction is the same, then the mass of water vapor formed as a product is:
mass reactants = mass products
mass of Hydrogen + mass of Oxygen = mass of water vapor
0.1 g + 0.8 g = 0.90 g
Or we can also solve<em> by using stochiometry</em> (using the concept of moles) to find the mass of water vapor
Answer:
Remaining the same
Explanation:
By the Lavoisier's principle the matter can't be created nor destroyed, but always transformed.
It means that in an ecosystem, the matter, and also the energy, is not increasing and not decreasing, the total amount remains the same, but in different forms.
Answer:
1. First one is true : as per periodic table down the group , the elements has increasing order of shell & with that the London dispersion forces brings the inter-molecules close together and bromine converted into liquid .
2. second one is False because carbon-carbon bonds are not weak bonds they form mutual covalent bonds which are stronger bonds and cannot be easily disrupted .
3. A single carbon atom has the valency of 4 so it can be bonded with four hydrogen atom at the same time .
Explanation:
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.