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3 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Chemistry

2 answers:

svet-max [94.6K]3 years ago

8 0

Explanation:

J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.

irinina [24]3 years ago

7 0

QUICK ANSWER

J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.

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Draw the structure for cis-2,3-dibromo-2-hexene.

padilas [110]

First, draw the 2-hexene. Th is is a molecule of six carbons with a double bond in the second carbon:

      CH3 - CH = CH2 - CH2 - CH2 - CH3

Secong, put one Br on the second carbon and one Br on the third carbon:

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Third, cis means that the two Br are placed in opposed positions, this is drawn with one Br up and the other down. So, you need to represent the position of the Br in the space:

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       |    |           |       |    |
H - C - C = C - C - C - C - H
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4 years ago

Read 2 more answers

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

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3 years ago

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Answer:

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Explanation:

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