Answer:
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
Explanation:
Step 1: Data given
the contribution of protons from H2SO4 is near 100 %.
Step 2: The unbalanced equation
.Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + H2O(l)
Step 3: Balancing the equation
On the left side we have 4x H (2x in Ba(OH)2 and 2x in H2SO4). On the right side, we have 2x H (in H2O).
To balance the amount of H on both sides, we have to muliply H2O (on the right side) by 2.
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
Step 4: The net ionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
After canceling those spectator ions in both side, look like this:
Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
Answer:
B.) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom loses the electron.
Explanation:
Silicon is located in the 3rd row and 14th column in the periodic table. Arsenic is located in the 4th row and 15th column in the periodic table. This means that arsenic has one more valence electron than silicon. Since arsenic is located one row down from silicon, its valence electrons occupy higher energy orbitals.
Silicon maintains a crystal-like lattice structure. Each silicon atom is covalently connected to assume this shape. When silicon gains one extra electron from arsenic, it experiences n-type doping. This new electron is not tightly bound in the lattice structure. This allows it to move more freely and conduct more electricity. This can also be explained using band gaps. Silicon, which previously had an empty conduction band, now has one electron in this band. This lowers the band gap between the conduction and valence bands and increases conductivity.
Both products will start to cancel the acidity and how strong the base is if they are mixed. If the acid is stronger than the base then it will be an acidic product and visa versa if the base is stronger than the acid.
Positron emission = emission of a positron and a neutrino when a
proton is convert into a neutron. The total number of particles in the
nucleus doesn't change, -1 proton +1 neutron
It's a spontaneous reaction for some nucleus.
eg:
Positron = e+
Neutrino=ve
O-15 --> N-15 + e+ +ve
Electron
capture= A nucleus absorb an electron while a proton is convert in a
neutron and emit a neutrino. The total number of particles in the
nucleus doesn't change, -1 proton +1 neutron
eg:
Al-26 +e- --> Mg-26 + ve
Electron
capture and positron emission are two mechanisms to explain the decay
of some unstable isotopes. Electron capture is usually observed when the
energy difference between the initial and final state is low. Mainly
because of the larger amount of kinetic energy need for the expulsion
two particles with the positron emission mechanism.
