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Mariulka [41]
2 years ago
5

Atoms of different chemical elements can be differentiated based on the number of.

Chemistry
1 answer:
Alona [7]2 years ago
3 0

The atoms of different chemical elements can be differentiated based on the number of protons.

Each chemical element is differentiated by his number of protons. This number of protons is also knowing as atomic number and it is represented by the letter (Z).

The atomic number represented the number of protons in the nucleus of an element. For example:

  • The element that has only 1 proton in his nucleus is Hydrogen atom.
  • The element that has 2 protons in his nucleus is Helium atom.

<h3>What is an atom?</h3>

The atom is the smallest part of the composition of matter, it is indivisible and is composed of a nucleus that has protons and neutrons, and around the nucleus there are the electrons.

Learn more about the atom at: brainly.com/question/17545314

#SPJ4

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3 years ago
How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2
jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

For the given chemical reaction:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

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