Answer:
Thomson placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. The results of these experiments helped Thomson determine the mass-to-charge ratio of the cathode ray particles, which led to a fascinating discovery, minus the mass of each particle was much, much smaller than that of any known atom. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what cathode material they originated from. From this evidence, Thomson made the following conclusions:
The cathode ray is composed of negatively-charged particles.
The particles must exist as part of the atom, since the mass of each particle is only ~1/2000 the mass of a hydrogen atom.
These subatomic particles can be found within atoms of all elements.
While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of the electron disproved the part of Dalton's atomic theory that assumed atoms were indivisible. In order to account for the existence of the electrons, an entirely new atomic model was needed.
Explanation:
An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. The dissolved electrolyte separates into cations and anions, which disperse uniformly through the solvent. Electrically, such a solution is neutral.
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The compound that is formed when a copper cation and a bromine anion reacts would be Copper (II) bromide. It will have a chemical formula CuBr2. This reaction is a synthesis reaction where two substances reacts to form one product.
Answer:
<h2>8.5 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>8.5 g/cm³</h3>
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Answer:
it cannot !!
Explanation:
a chemical reaction MUST occur to separate a compound! :)
