Answer:
Yea.....where is the article and the 6 questions?
0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure
0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8
The presence of unsaturation that is carbon-carbon double or triple bonds can be tested qualitatively by bromine test. In this test, the unknown sample is treated with a small amount of elemental bromine in organic solvent (which is deep brown in color). The disappearance of deep brown color of bromine results due to reaction of bromine with unsaturated sample.
The reaction of eugenol with bromine is given as:
Since the number of atoms of each element in the reactant side is equal to the number of atoms of elements of product side. Thus, the reaction is balanced.
The reaction between eugenol and bromine is shown in the image.
Since the percent yield of the reaction is 98%, that means that 1.7g of ammonia is 98% of the theoretical yield so you need to find what the theoretical yield is by dividing 1.7 by .98 to get 1.735g. Since you know that the theoretical yield is 1.735g you can use stoichiometry to figure out how much N₂ is required in order to produce 1.735g of ammonia.
To do that you first need to find the balanced chemical equation for the reaction which is N₂+3H₂⇒2NH₃. Then you need to find how many moles of ammonia are in 1.735g by dividing 1.735 by the molar mass of ammonia (17g/mol) to get 0.1021mol of ammonia. After that you can find the number of moles of N₂ required to make 0.1021mol of ammonia by using the fact that 1mol N₂ is used to make 2mol NH₃ so you divide 0.1021 by 2 to get 0.05103mol N₂. lastly you can find the number of grams of N₂ by multiplying 0.05103 by the molar mass of N₂ (28g/mol) to get 1.42g of N₂. Therefore 1.4g of N₂ is required to make 1.7g of ammonia.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
Explanation:
Going across a period, the effective nuclear charge increase (b/c you are at the same energy level "n" and you are adding more protons) so the electrons are harder to remove and the ionization energy increases. Shielding does not really affect the nuclear pull across the period b/c you are not going farther away from nucleus as you move across a period (you move farther away when you move down a group) you are staying in the same period just adding more protons which increases the pull on electrons and making it harder to remove electrons thus ionization energy increases.