<u>Answer:</u> The net ionic equation contains
ions
<u>Explanation:</u>
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of ammonium sulfate and calcium nitrate is given as:
![(NH_4)_2SO_4(aq.)+Ca(NO_3)_2(aq.)\rightarrow CaSO_4(s)2NH_4NO_3(aq.)](https://tex.z-dn.net/?f=%28NH_4%29_2SO_4%28aq.%29%2BCa%28NO_3%29_2%28aq.%29%5Crightarrow%20CaSO_4%28s%292NH_4NO_3%28aq.%29)
Ionic form of the above equation follows:
![2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ca^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow CaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)](https://tex.z-dn.net/?f=2NH_4%5E%7B%2B%7D%28aq.%29%2BSO_4%5E%7B2-%7D%28aq.%29%2BCa%5E%7B2%2B%7D%28aq.%29%2B2NO_3%5E%7B-%7D%28aq.%29%5Crightarrow%20CaSO_4%28s%29%2B2NH_4%5E%2B%28aq.%29%2B2NO_3%5E-%28aq.%29)
As, ammonia and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
![Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)](https://tex.z-dn.net/?f=Ca%5E%7B2%2B%7D%28aq.%29%2BSO_4%5E%7B2-%7D%28aq.%29%5Crightarrow%20CaSO_4%28s%29)
Hence, the net ionic equation contains
ions
Answer:
it is used for welding and cutting
Halogens (atoms with 7 valence electrons) and Hydrogen
or generally, atoms with their shells almost full
A= 6
B= 2
C= 1
D= 5
E= 3
F= 4
<h3>
Answer:</h3>
1.43 × 10⁻²⁰ mol Li
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
8.63 × 10³ atoms Li
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})](https://tex.z-dn.net/?f=%5Cdisplaystyle%208.63%20%5Ccdot%2010%5E3%20%5C%20atoms%20%5C%20Li%28%5Cfrac%7B1%20%5C%20mol%20%5C%20Li%7D%7B6.022%20%5Ccdot%2010%5E%7B23%7D%20%5C%20atoms%20%5C%20Li%7D%29)
- Multiply/Divide:
![\displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.43355%20%5Ccdot%2010%5E%7B-20%7D%20%5C%20moles%20%5C%20Li)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li