Halogen are the most reactive due to their electronic configuration
Answer:
8.547 x 10⁴disintegrations per second
Explanation:
To calculate the disintegrations per second as -
Given ,
2.31 μCi of sulfur -35 .
Since ,
1 Ci = 3.7 * 10 ¹⁰ Bq
1 μCi = 10 ⁻⁶ Ci
Hence ,
conversation is done as follows -
2.31 ( 1 * 10⁻⁶) * ( 3.7 * 10¹⁰)
= 8.547 x 10⁴
Hence ,
8.547 x 10⁴disintegrations per second , the sample undergo for it to be brand new .
Atmospheric
pressure<span>, sometimes also called barometric pressure, is the pressure exerted by the weight of air in
the </span>atmosphere of Earth<span> (or that of another planet)</span>
1 atm is equivalent to = 101325
Pa
= 760 mmHg
= 760 torr
= 1.01325 bar
So 1.23 atm is equal to
= 124629.8 Pa
= 934.8 mmHg
= 934.8 torr
<span>= 1.2462 bar</span>
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
Potassium oxide: K₂O.
There's no need for prefixes since K₂O is an ionic compound.
<h3>Explanation</h3>
Find the two elements on a periodic table:
- Potassium- K- on the left end of period four.
- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
- Oxygen is a nonmetal.
A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:
- Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
- Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion
when it combines with metals.
The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each ion is twice that on a 
ion. Each would pair up with two . Hence the subscript in the formula: 
.
There are two classes of compounds:
- Covalent compounds, which need prefixes, and
- Ionic compounds, which need no prefix.
Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide 
, the prefix di- indicates that there are two oxygen atoms in the formula . However, there's no need for prefix in ionic compounds such as 
.
