Answer:
3) The calculated density would be high, because the volume would be incorrectly measured low.
Explanation:
Density is defined as mass per unit volume. It means the density is inversely proportional to the volume.
A solid with a hollow center will have high volume due to air inside the hollow area that will lead to low density.
But the error in the calculation of density is that the "density would be high, because the volume would be incorrectly measured low."
Hence, the correct answer is "3)"
Answer:
Antoine Lavoisier and Johann Wolfang Döbereiner organized the elements based on properties such as how the elements reacts or whether they are solid or liquid.
Explanation:
The periodic table of the elements as we have it today was developed as a result of the work of several notable centuries who lived centuries apart, all of who made notable contributions to development of the modern periodic table in use today.
In 1789, Antoine Lavoisier, a French Chemist provided a definition of elemets which he defined as a substance whose smallest units cannot be broken down into a simpler substance. He further grouped the elements into two as metals and nonmetals.
In 1829, German physicist Johann Wolfang Döbereiner arranged elements in groups of three in increasing order of atomic weight and called them triads. His arrangement owasf elements into triads was based on his observation of similarities in physical and chemical properties of certain elements.
John Newlands, a British Chemist was the first to arrange the elements into a periodic table with increasing order of atomic masses.
In 1869, Russian chemist Dmitri Mendeleev developed a periodic table which provided a framework the modern periodic table. He arranged the elements according to their atomic weight, leaving gaps for elements that were yet to be discovered.
The modern periodic table arranges elements based on increasing atomic number.
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
A starfish has five equal arms, so a starfish would have 5 lines of symmetry. Hope this helps!
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V