Answer:
Ionization energy increases going left to right across a period and increases from bottom to top in a group
Electron affinity increases when going up a group
If we are excluding noble gases (aka group 8/18), Chlorine is the element that has the greatest electron affinity. This is because Fluorine's 2p orbital is limited and packed which doesn't quite allow sharing of the orbital with extra electrons easily, while Chlorine has a 3p orbital allowing more space for electrons, where the orbital electrons would be inclined to do so.
Helium is the element with the greatest ionization energy since it's at the top and energy (from Oganesson to Helium) increases when going across a period (from Hydrogen to Helium).
Answer: journals
explanation: A publication is defined as anything that has been published: a book. a research paper, a news article. A research paper describes the output of research. If it is published in a Journal or conference, then it is a published research paper, or a publication.
Answer:
The limiting reactant is the 6.279 g of ![NiCl_2](https://tex.z-dn.net/?f=NiCl_2)
Explanation:
We have to start with the <u>reaction</u> between sodium carbonate (
) and the Nickel (II) Chloride (
), so:
![Na_2CO_3~+~NiCl_2-->~NiCO_3~+~NaCl](https://tex.z-dn.net/?f=Na_2CO_3~%2B~NiCl_2--%3E~NiCO_3~%2B~NaCl)
We will have a <u>double replacement reaction</u>. Now we have to <u>balance</u> the reaction, so:
![Na_2CO_3~+~NiCl_2-->~NiCO_3~+~2NaCl](https://tex.z-dn.net/?f=Na_2CO_3~%2B~NiCl_2--%3E~NiCO_3~%2B~2NaCl)
The next step is the <u>calculation of the moles for each reactive</u>. For
we have use the <u>molarity equation</u>:
![M~=~\frac{mol}{L}](https://tex.z-dn.net/?f=M~%3D~%5Cfrac%7Bmol%7D%7BL%7D)
![0.1010~M~=\frac{mol}{0.5~L}](https://tex.z-dn.net/?f=0.1010~M~%3D%5Cfrac%7Bmol%7D%7B0.5~L%7D)
![mol~=~0.1010*0.5=~0.0505~mol~of~Na_2CO_3](https://tex.z-dn.net/?f=mol~%3D~0.1010%2A0.5%3D~0.0505~mol~of~Na_2CO_3)
For the calculation of moles of
we have to use the <u>molar mass</u> of the compound (129.59 g/mol):
![6.279~g~NiCl_2\frac{1~mol~NiCl_2}{129.59~g~NiCl_2}=~0.0484~mol~NiCl_2](https://tex.z-dn.net/?f=6.279~g~NiCl_2%5Cfrac%7B1~mol~NiCl_2%7D%7B129.59~g~NiCl_2%7D%3D~0.0484~mol~NiCl_2)
The next step is the division of each mole value by the <u>coefficient of each reactive</u> of the balance reaction. In this case <u>we have "1" for each reactive</u>, so:
![\frac{0.0484}{1}=0.0484](https://tex.z-dn.net/?f=%5Cfrac%7B0.0484%7D%7B1%7D%3D0.0484)
![\frac{0.0505}{1}=0.0505](https://tex.z-dn.net/?f=%5Cfrac%7B0.0505%7D%7B1%7D%3D0.0505)
The final step is to <u>choose the smallest value</u>. In this case is the value that correspond to
. Therefore
is the limiting reactive.