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3 years ago

From his experiments ,j.j Thomson concludes what?

Chemistry

2 answers:

Vladimir [108]3 years ago

7 0

Answer:

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 ∼ 20001start fraction, 1, divided by, 2000, end fraction the mass of a hydrogen atom.

Explanation:

Andrews [41]3 years ago

4 0

Answer:

J J Thomson's experiment confirms the existence of corpuscles (electrons), by concluding that: cathode rays are charges of negative electricity carried by particles of matter.

Explanation:

J J Thomson's experiment with cathode ray tubes helped him to discover the electron (which Dalton did not know about).

Cathode ray is composed of negatively-charged particles. And these particles must exist as part of the atom, since the mass of each particle is only ∼ 20001start fraction, 1, divided by, 2000, end fraction the mass of a hydrogen atom. And also, that the particles that made up the gases were universal.

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3 years ago

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Answer:

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Explanation:

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3 years ago

Use Avogadro's number, 6.02E23, to calculate the number

Zepler [3.9K]

Answer:

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3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

irinina [24]

Answer:

$m_{KBr}=6.030gKBr$

Explanation:

Hello.

In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

$(T_b-T_b_0)=i*m*K_b$

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

$m=\frac{T_b-T_b_0}{i*K_b}=\frac{(119.3-117.8)\°C}{2*1.48\°C*kg*mol^{-1}}\\ \\m=0.507mol/kg$

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

$n_{KBr}=0.507mol/kg*0.1kg=0.0507mol$

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

$m_{KBr}=0.0507mol*\frac{119g}{1mol} \\\\m_{KBr}=6.030gKBr$

Best regards.

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3 years ago