The answer is letter a. Hill terrain, hill slope, and hill elevation
A hydrogen bond<span> is the electrostatic attraction between two polar groups that occurs when a </span>hydrogen<span> (H) atom covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative atom nearby. examples h20</span>
LiBr.
<h3>Explanation</h3>
Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).
Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be
.
Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be ![\text{Br}^{-}](https://tex.z-dn.net/?f=%5Ctext%7BBr%7D%5E%7B-%7D)
All the ions in an ionic compound carry charge. However, some of the ions like
are positive. Others ions like
are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.
1 × (+1) + 1 × (-1) = 0. The positive charge on one
ion balances the negative charge on one
ion. The two ions would pair up at a 1:1 ratio.
The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions.
is positive and
is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.
Mass of Oxygen: 0.0159 grams
Moles of Oxygen: 9.94x10^-4
To find the mass of oxygen, subtract the mass of copper from the total mass.
![0.143-0.1271=0.0159](https://tex.z-dn.net/?f=0.143-0.1271%3D0.0159)
There are 0.0159 grams of Oxygen.
To find how many moles there are, divide the given amount of oxygen by the molar mass (atomic mass) of oxygen because that mass is the same as one mole of oxygen.
Molar mass of Oxygen: 16.00
![0.0159/16.00=9.94*10^{-4}](https://tex.z-dn.net/?f=0.0159%2F16.00%3D9.94%2A10%5E%7B-4%7D)
There are 9.94*10^-4 moles of Oxygen.