Answer:
In first shell only 2 electrons are present in hydrogen only one electron is present in valence shell. To complete its duplet hydrogen can share lose or gain only one electron to form chemical bond.
Hydrogen can form only one bond.
Explanation:
Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.
<h3>What is mass?</h3>
The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.
Given,
Molar mass of barium nitrate = 261.35 g/mol
If,
have a mass of 261.35 g/mol then,
formula units will have a mass of,
![\begin{aligned}& = \dfrac{261.35 \times 5.30 \times 10^{22}}{6.022\times 10^{23}}\\\\&= 23.0\;\rm gm\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20%3D%20%5Cdfrac%7B261.35%20%5Ctimes%205.30%20%5Ctimes%2010%5E%7B22%7D%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%7D%5C%5C%5C%5C%26%3D%2023.0%5C%3B%5Crm%20gm%5Cend%7Baligned%7D)
Therefore, option C. 23.0 gm is the mass of barium nitrate.
Learn more about mass here:
brainly.com/question/24958554
Answer:
A linear pattern is a continuous decrease or increase in numbers over time. On a graph, this data appears as a straight line angled diagonally up or down (the angle may be steep or shallow). So the trend either can be upward or downward.
Explanation:
The half-life of carbon-14 is about 5730 years
Answer:
A. 1:3
Explanation:
If we look at the ions shown in the image attached to the question, we will notice that we have aluminum (Al^3+), a trivalent ion combining with the iodide ion (I^-).
Aluminum can easily give out its three outermost electrons to three atoms of iodine. If aluminum gives out its three electrons, it achieves the stable octet structure. Iodine atoms have seven electrons in their outermost shell. They only need one more electrons to complete their octet. This one electron can be gotten by the combination of three iodine atoms with one atom of aluminum. One electron each is transferred from the aluminum atom to each iodine atom to form AlI3 with a ratio of 1:3.