Answer:
See attachment.
Explanation:
Elements that are in the same group will definitely possess similar characteristics because they tend to have the same valence electron which determines their reactivity.
On a periodic table, elements in the same group can be found arranged on the same column in the periodic table.
Therefore the two elements that have similar characteristics are those two elements you can see on the same column in group 2. See the two elements indicated in the attachment below.
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
The pH of the buffer can be known as
The concentration of
Similarly, the concentration of [HA] =
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]
So, the pH of the buffer is 6.1236.
The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.