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3 years ago

Whose model of the atom did J.J. Thomson prove to be incorrect with his Cathode-Ray tube experiments?

Chemistry

2 answers:

madreJ [45]3 years ago

8 0

Answer:

John Dalton

Explanation:

J.J. Thompson’s cathode ray experiment proved that Dalton’s atomic theory was incorrect,

Dalton proposed that atoms are indivisible and they are the smallest part of a substance. But Thompson’s cathode ray experiment showed that atoms are made up of negatively charged particles. Option B is correct.

Option A is incorrect. Neils Bohr gave the atomic model after J.J. Thompson and he explained the quantization of electron energy.

Option C is incorrect. Erwin Schrodinger proposed that electrons move in electron clouds, electrons do not have a definite position and there is a probability of finding electron in space.

Option D is incorrect. James Chadwick discovered the neutron.

Brut [27]3 years ago

7 0

JJ Thomson proved John Dalton's theory wrong, as Thomson discovered the electron. This showed that there were particles that were smaller than an atom, despite Dalton claiming that the atom is the smallest unit of matter.

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3 years ago

The use of phenol (carbolic acid) as a wound disinfectant was first practiced by:

uranmaximum [27]

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Explanation:

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2 years ago

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kolezko [41]

Answer:

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2 years ago

Read 2 more answers

Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to: