answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mash [69]
2 years ago
9

Imagine if during the cathode ray experiment, the size of the particles of the ray was the same as the size of the atom forming

the cathode. Which other model or scientific observation would have also been supported?
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
This would support Bohr's prediction about electrons moving in orbits having specific energy.
This would support Bohr's prediction about electrons being randomly scattered around the nucleus in the atom.
This would support Dalton's postulates that proposed that atoms combine in fixed whole number ratios to form compounds.
Chemistry
2 answers:
Yakvenalex [24]2 years ago
7 0

Answer:

This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.

Explanation:

Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.

Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.

If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.

torisob [31]2 years ago
4 0

Answer:

This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.

Explanation:

Dalton proposed that all matter is made up of atoms. Atoms are indivisible. But Thompson’s cathode ray experiment proved that the cathode rays are made up of negatively charged particles which were deflected by both electric and magnetic field. This caused Thompson to propose the Plum and Pudding model of the atom.

You might be interested in
If a sample containing 2.50 ml of nitroglycerin (density=1.592g/ml) is detonated, how many total moles of gas are produced?
masya89 [10]
<span>0.127 moles The formula for nitroglycerin is C3H5N3O9 so let's first calculate the molar mass of it. Carbon = 12.0107 Nitrogen = 14.0067 Hydrogen = 1.00794 Oxygen = 15.999 C3H5N3O9 = 3 * 12.0107 + 5 * 1.00794 + 3 * 14.0067 + 9 * 15.999 = 227.0829 Now calculate the number of moles of nitroglycerin you have by dividing the mass by the molar mass 2.50 ml * 1.592 g/ml / 227.0829 g/mol = 0.017527 mol The balanced formula for when nitroglycerin explodes is 4 C3H5N3O9 => 12 CO2 + 10 H2O + O2 + 6 N2 Since all of the products are gasses at the time of the explosion, there is a total of 29 moles of gas produced for every 4 moles of nitroglycerin Now multiply the number of moles of nitroglycerin by 29/4 0.017527 mol * 29/4 = 0.12707075 moles Round to 3 significant figures, giving 0.127 moles</span>
3 0
2 years ago
What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of car
Basile [38]

The reaction between boron sulfide and carbon is given as:

2B2S3 + 3C → 4B + 3CS2

As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.

Given data:

Mass of C = 2.1 * 10^ 4 g

Mass of B = 3.11*10^4 g

Mass of CS2 = 1.47*10^5

Mass of B2S3 = ?

Now based on the law of conservation of mass:

Mass of B2S3 + mass C = mass of B + mass of CS2

Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5

Mass of B2S3 = 15.7 * 10^4 g


4 0
2 years ago
What other objects could be used to simulate radioactive and nonradioactive nuclei? Check all that apply.
Ray Of Light [21]

Answer:

quarters

a computer that shows pictures of atoms on screen

candy with letters on one side

Explanation:

3 0
2 years ago
How many moles of nitrogen are in 3.7 moles of C8H11NO2?
Phoenix [80]
<h3>Answer:</h3>

               3.7 Moles of Nitrogen

<h3>Explanation:</h3>

                      On observing the chemical formula C₈H₁₁NO₂ (might be formula of Dopamine) it is found that one mole of this compound contains;

8 Moles of Carbon

11 Moles of hydrogen

1 Mole of Nitrogen and

2 Moles of Oxygen respectively.

<u>Calculate Number of Moles of Nitrogen:</u>

As,

                   1 Mole of C₈H₁₁NO₂ contains  =  1 Mole of Nitrogen

So,

            3.7 Moles of C₈H₁₁NO₂ will contain  =  X Moles of Nitrogen

Solving for X,

                       X  =  (3.7 Moles × 1 Mole) ÷ 1 Mole

                       X  =  3.7 Moles of Nitrogen

4 0
2 years ago
Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O
mestny [16]

86.5\; \% of octane had been converted to carbon dioxide CO₂.

<h3>Explanation</h3>

Octane has a molar mass of

12.01 \times 8 + 1.008 \times 18 = 114.22 \; \text{g} \cdot \text{mol}^{-1}

1.000 gallon of this fuel would have a mass of 2.650 kilograms or 2.65 \times 10^{3} \; \text{g}, which corresponds to 2.65 \times 10^{3} / 114.22 = 23.2\; \text{mol} of octane.

Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

2\; \text{C}_8\text{H}_{18} + 25 \; \text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

2\; \text{C}_8\text{H}_{18} + 17 \; \text{O}_2 \to 16 \; \text{CO} + 18 \; \text{H}_2\text{O}

The mass of the product mixture is 11.53 - 2.65 = 8.88 \; \text{kg} heavier than that of the octane supplied. Thus 8.88 \; \text{kg} = 8.88 \times 10^{3} \; \text{g} of oxygen were consumed in the combustion. There are 277.5 \; \text{mol} of oxygen molecules in 8.88 \times 10^{3} \; \text{g} of oxygen.

Let the number of moles of octane that had undergone complete combustion as seen in the first equation be x (0 \le x \le 23.2). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal 23.2 - x.

25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.

n(\text{O}_2, \; \text{Complete Combustion}) + n(\text{O}_2, \; \text{Incomplete Combustion} ) = n(\text{O}_2, \; \text{Consumed})\\

\frac{25}{2} \; x + \frac{17}{2} \; (23.2 - x) = 277.5\\4 \; x = 277.5 - \frac{17}{2} \times 23.2\\x = 20.1

Therefore 20.1 \; \text{mol} out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.

\%n(\text{Complete Combustion}) = 20.1 / 23.2 \times 100 \; \%= 86.5\; \%

7 0
2 years ago
Other questions:
  • What is four thing the atmosphere does for us
    11·2 answers
  • Which of the following does not describe elements?
    8·2 answers
  • A scientist has isolated a chemical compound that is partially soluble in water. The chemical helps keep water-soluble substance
    14·1 answer
  • How many liters of a 0.352 M solution of Ca(SO4) would contain 62.1 g of Ca(SO4)?​
    7·1 answer
  • Why is useful to use moles to measure chemical quantities
    14·1 answer
  • Platinum, which is widely used as a catalyst, has a work function φ(the minimum energy needed to eject an electron from the meta
    8·1 answer
  • Mrs. Jacobson is pushing a fridge to the right with a force of 100 N. The force pushing it
    8·2 answers
  • Predict what will observe in below mention experiment.
    10·1 answer
  • 4. What are the potential sources of error that might cause disagreement between the activity series' prediction of reactions an
    7·1 answer
  • Calculate the mass of 3.75 x 10^23 molecules of CaSO4 Please HelpIn a hurry Ill give brainliest
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!