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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.

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vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂  + 4KI    →    2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

                        Cu(NO₃)₂       :              KI    

                              2              :               4

                            0.013          :            4 × 0.013=0.052 mol

Volume of KI:

<em>Molarity = moles of solute / volume in L</em>

volume in L = moles of solute /Molarity

volume in L =  0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0
2 years ago
Is utensils a substance homogeneous mixture or heterogeneous mixture
Phoenix [80]

A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

7 0
2 years ago
Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume
AnnZ [28]

Answer : The correct options are,

(B) V\propto \frac{1}{P}

(C) P\propto \frac{1}{V}

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

V\propto \frac{1}{P}

The relation between the pressure and volume of two gases are:

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

5 0
2 years ago
Explain why you hear a “whoosh” sound when you open a can containing a carbonated drink. Which gas law applies?
Lana71 [14]

Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).

6 0
2 years ago
Read 2 more answers
A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na
notka56 [123]

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = \frac{137.327g/mol}{233.38g/mol}\times 0.4105g of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100

% mass of Barium ions = 35.72%.

8 0
2 years ago
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