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mariarad [96]
4 years ago
6

What did John Dalton contribute to the understanding of the atom?

Chemistry
1 answer:
vodka [1.7K]4 years ago
6 0
Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.
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How do movements and interactions of air masses cause change in the weather?
ollegr [7]
The way the air moves affects the weather, because winds move heat and cold temperatures as well as moisture from one place to another, transporting conditions from one geographical zone to another. The way winds pass each other, and the direction they move, also affects what weather a region will see on any given day.
5 0
3 years ago
A ground test utilizing an auxiliary current electrode and an auxiliary potential electrode is known as the______.
kompoz [17]

Answer:

The three-point test

Explanation:

The three-point test refers to a ground test utilizing an auxiliary current electrode and an auxiliary potential electrode.

7 0
2 years ago
What is the overall reaction order for this rate law: rate= k[a][b]?
SpyIntel [72]
The overall reaction order is 2, because you add the orders of the individual components, which each are 1. 
7 0
3 years ago
Two well-known complex ions containing ni are [ni(h2o)6]2+, which is green, and [ni(en)3]2+, which is purple. which one of these
Julli [10]
For [Ni(en)³]²⁺ which is purple, the crystal field splitting energy is greater than the complex ion, [Ni(H₂O)₆]²⁺ which is green in color.
When a Lewis base id attached to the metal ion by covalent bond, then the complex ion is formed and when these complex ions are present with other ions of opposite charge or neutral charge, they will make complex compounds.
7 0
3 years ago
if .40 L if water is added to the volume of cup 3, what would be the new molarity of a 2 M solution of kool-aid
stiks02 [169]

Answer:

0.7692 M ≅ 0.77 M.

Explanation:

  • It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.
  • It can be expressed as:

<em>(MV) before dilution = (MV) after dilution.</em>

M before dilution = 2.0 M, V before dilution = 0.25 L.

M after dilution = ??? M, V after dilution = 0.25 L + 0.40 L = 0.65 L.

∴<em> M after dilution = (MV) before dilution/(V) after dilution</em> = (2.0 M)(0.25 L)/(0.65 L) =<em> 0.7692 M ≅ 0.77 M.</em>

8 0
3 years ago
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