Answer:
Spanish
Explanation:
help is urgent the log of wood and the same
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
Answer:
This question is incomplete but the correct option is B
Explanation:
This question is incomplete because of the absence of the "Reference Table S", however the question can still be answered in the absence of the table. The energy described in the question is the ionization energy (energy required to remove the most loosely bound electron in an atom). This question seeks to know the atom (from the options provided) with the least ionization energy.
Ionization energy increases from left to right across the period because it's easier to remove a single electron (valence electron) from the outermost shell than to remove two electrons from the same shell; thus the more the valence electrons (in a shell), the higher the ionization energy. Thus, bromine (Br) and tin (Sn) have high ionization energies because they have more number of electrons in there outermost shell.
<u>Berylium (Be) and strontium (Sr) are both in the group 2 of the periodic table because they both have 2 electrons in there outermost shell. Ionization energy decreases down a group. This is because the farther an electron is from the nucleus, the weaker the force of attraction between the nucleus and the electron. Thus, strontium (Sr) would have a lesser ionization energy between the two and would indeed have the least ionization among the options provided</u>. Hence, the correct option is B
Answer:
The advantage of this technique is that purified water as well as deposited metals can be re-used. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode.
