Answer:- ![[Na_2CO_3]=0.254M](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D0.254M)
, ![[Na^+]=0.508 M](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D0.508%20M)
, ![[CO_3^2^-]=0.254M](https://tex.z-dn.net/?f=%5BCO_3%5E2%5E-%5D%3D0.254M)
Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.
Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL. Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.
Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:
= 
So, molarity of sodium carbonate solution is 0.254 M.
sodium carbonate dissociate to give the ions as:
There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.
![[Na^+]=2(0.254M)](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D2%280.254M%29)
= 0.508 M
There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.
I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
Answer:
The organs present inside the chest are :
1. The lungs
2. The heart
Explanation:
The chest cavity is also called as the thoracic cavity. It is the second largest hollow space of the body.In the bottom , it is enclosed by the diaphragm.
This cavity actually contain three space each round with mesothelium , pleural cavity and precardial cavity.
This contain the lungs , the tracheobronchial tree , the heart , the blood vessels which transport the blood between the heart and the lungs.
It also contain the esophagus .
Esophagus is the path through which the food passes from the mouth to the stomach.
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
