The given question is incomplete. the complete question is:
The world burns the fossil fuel equivalent of approximately 
kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)
Answer: 
Explanation:
Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
To calculate the moles :
According to stoichiometry :
As 2 moles of octane give = 16 moles of 
Thus 
of octane give =
of
Mass of 
Thus of is produced by world fossil fuel combustion per year.
Answer: voltage drops in each resistor ΔU= RI
Explanation: if lamps or other resistor which cause load are in series in
Electric circuit, current I passing circuit is same. Voltage decreases
In every resistor
The complete answer contains the answer choices.
This is the complete question:
Atoms of which element react spontaneously with Mg2+(aq)? (1)chromium
(2)barium
(3)iron
(4)zinc
Answer: option 2. barium.
Explanation.
1) Mg ²⁺ (aq) is the aquous cation of the metal Mg.
2) Only a metal more reactive than Mg will be able to exchange with it the oxidation states.
3) Mg is an earth alkalyne metal (group 2 of the periodic table). The metals from this group are more reactive than the metals in the groups to its right: 3, 4, 5, 6, 7, ... Only the alkalyne metals (those in group 1) are more reactive than the earth alkalyne metals.
4) Of the list, barium is the only alkalyne metal, so it is more reactive than Mg and will be able to deliver 2 electrons to transform the cations Mg²⁺ into Mg while the very Ba will become Ba²⁺.
5) Chromium, iron and zinc are transition metals, so less metallic (reactive) than Mg.
Answer:
pent-3-ene-1-yne
Explanation:
1 2 3 4 5
CH ≡ C - CH = CH - CH3
IUPAC name : Pent-3-ene-1-yne
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.
