The IUPAC name for the given product is 2 chloro Butane.
<h3>What is IUPAC nomenclature?</h3>
IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.
- In the given product total four carbon atoms are present and between all of them single bonds are present.
- In the second carbon atom, chlorine group is present.
- During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.
Hence name of the product is 2 chloro Butane.
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Answer: The IUPAC name of 
is 5-chloro-2-pentyne
Explanation:
1. First select the longest possible carbon chain. For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
2. The longest possible carbon chain should contain all the bonds and functional groups.
3. The numbering is done in such a way that the carbon containing the functional group or substituent gets the lowest number. Triple bond is given priority over substituent halogen.
4. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne.
Thus the IUPAC name of is 5-chloro-2-pentyne
the answer is d. this is due to the fact a proton weighs 2000 times more then a electron
The number of moles of NH3 that could be made would be 0.5 moles
<h3>Stoichiometric reactions</h3>
From the balanced equation of the reaction:
N2 (g) + 3 H2(g) ----> 2NH3 (g)
The mole ratio of N2 to H2 is 1:3
Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.
Mole ratio of H2 and NH3 = 3:2
Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:
2 x 0.75/3 = 0.5 moles
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Answer:
The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.
Explanation:
The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.
From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.
The individual balanced equation of reaction is:
C3H3 +5O2 >> 3CO2 + 4H2O
and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.
The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.
