Answer:
Obtención. El carbono se encuentra - frecuentemente muy puro - en la naturaleza, en estado elemental, en las formas alotrópicas diamante y grafito. El material natural más rico en carbono es el carbón (del cual existen algunas variedades). Grafito: Se encuentra en algunos yacimientos naturales muy puro.
Answer:
The molarity of the dissolved NaCl is 6.93 M
Explanation:
Step 1: Data given
Mass of NaCl = 100.0 grams
Volume of water = 100.0 mL = 0.1 L
Remaining mass NaCl = 59.5 grams
Molar mass NaCl= 58.44 g/mol
Step 2: Calculate the dissolved mass of NaCl
100 - 59. 5 = 40.5 grams
Step 3: Calculate moles
Moles NaCl = 40.5 grams / 58.44 g/mol
Moles NaCl = 0.693 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity dissolved NaCl = 0.693 moles / 0.1 L
Molarity dissolved NaCl = 6.93 M
The molarity of the dissolved NaCl is 6.93 M
I found these four statements for that question:
Each molecule contains four different elements.
Each molecule contains three atoms.
Each molecule contains seven different bonds.
Each molecule contains six oxygen atoms.
The last one is true. Each molecule contains six oxygen atoms.
The number to the right of O and of (NO3) ares subscripts.
The chemical formula uses subscripts to indicate the number of atoms.
The subscript 2 in (NO3)2 means that there are two NO3 radicals.
And the subscript 3 to the right of O means that each NO3 radical has three atoms of O.
Then, the number of atoms of O is 2 * 3 = 6.
So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.
From that molecule you can also tell:
- Each molecule contains one atom of barium
- Each molecule contains two atoms of nitrogen
- Each molecule contains two NO3 radicals
Answer:
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Explanation:
Answer:
, the solution is basic
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
Putting in the values:
![pOH=-\log[5.5\times 10^{-5}]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B5.5%5Ctimes%2010%5E%7B-5%7D%5D)



![9.8=-\log [H_3O^+]](https://tex.z-dn.net/?f=9.8%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![[H_3O^+]=1.6\times 10^{-10}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.6%5Ctimes%2010%5E%7B-10%7DM)
As pH is more than 7, the solution is basic