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timofeeve [1]
4 years ago
12

J.J. Thomson discovered what part of the atom using a cathode ray tube (CRT?

Chemistry
1 answer:
kolbaska11 [484]4 years ago
4 0
J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged "soup."
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Explain how lines latitude and longitude help people find locations on earth.
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8 0
3 years ago
How many moles of oxygen are needed for the complete combustion of 12.6 moles of acetylene?
soldi70 [24.7K]

The moles of oxygen required to completely react with 1-mole acetylene is 2.5 mol.

The moles of reactant and product in a chemical reaction to the whole number ratio is given by the stoichiometric coefficient of the balanced chemical equation.

8 0
3 years ago
Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io
ryzh [129]
We are given with
Cobalt phosphate - CoPO4

We are asked for the net ionic equation for the phosphate dissolving in H3O+

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CoPO4 (s) + H3O+ (aq) ----->  HPO42- (aq) + Co3+ (aq) + H2O *(l)
8 0
3 years ago
Read 2 more answers
1. When you are underwater, sounds that you hear from above the water sound much quieter than when you are not underwater. Draw
GuDViN [60]

Answer:

Explanation:

the water make it sound different then above water

hop this helps

5 0
3 years ago
Determine the frequency and wavelength (in nm) of the light obsorded when the e- goes from n=1 to n=5
morpeh [17]

Answer:

Frequency = 3.16 ×10¹⁴ Hz

λ = 0.95×10² nm

Explanation:

Energy associated with nth state is,

En =  -13.6/n²

For n = 1

E₁ = -13.6 / 1²

E₁ = -13.6/1

E₁ = -13.6 ev

Kinetic energy of electron = -E₁ = 13.6 ev

For n = 5

E₅ = -13.6 / 5²

E₅ = -13.6/25

E₅ = -0.544 ev

Kinetic energy of electron = -E₅ = 0.544 ev

Wavelength of radiation emitted:

E = hc/λ = E₅ - E₁

hc/λ = E₅ - E₁

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.544ev  - (-13.6 ev  )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 20.9 ×10⁻¹⁹ J

λ = 0.95×10⁻⁷ m

m to nm:

0.95×10⁻⁷ m ×10⁹nm/1 m

0.95×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /0.95×10⁻⁷ m

Frequency = 3.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 3.16 ×10¹⁴ Hz

4 0
3 years ago
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