Answer:
First, you have to analyze your problem or question. After you research and collect data about your topic, create a hypothesis to test to try and find the answer. After testing your hypothesis, come up with a conclusion based on the results.
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
CuCl2+F2—>CuF2+Cl2.
This is a single replacement because there is one compound and one element. Picture Cu as ‘A’ Cl2 as ‘B’ and F2 as ‘C.’ So AB+C—>AC+B. A and B “broke up” and that resulted to A going with C to create the compound CuF2 leaving Cl2 alone.
This is your answer:
-13 degrees C
