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3 years ago

Describe the location of the positive charge of the atom in j. j. thomson’s plum pudding model.

Chemistry

2 answers:

Umnica [9.8K]3 years ago

5 0

Answer:

The positive charge of the atom in J.J. Thomson´s plum pudding model was all around like a cloud.

Explanation:

J.J Thomson´s plum pudding model quoted the atom was a positive cloud or a positive cake with electrons like balls in the middle of the cloud or cake, as you can see in the image below

katovenus [111]3 years ago

3 0

Thompson proposed a model (the plum pudding model) whereby the negatively charged corpuscles were distributed in a uniform sea of positive charge. In other words the positive charge was like the pudding and the negative charge were like raisins on the pudding.

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3) The spectrum of the unknown mixture is a mixture of two substances among the five. What are these two substances

Neporo4naja [7]

Answer:

a. H + d. Na

Explanation:

This is Mixture of H and Na. Because of H and in the emission and Absorption spectrum of H Total 5 lines [Red, Green, Blue and Purple (2)) are present and in the spectrum of Na Two yellow lines are present. In the mixture all this lines are are also 5 + Present. Total lines are 5+2= 7 (H+ Na)

Therefore, the answer is a. H + d. Na

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2 years ago

An outdoor grill has a propane (C3H8) tank that holds 4.8 L of gas at a pressure of 5.5 atm. If all of the gas is transferred in

shtirl [24]

Answer: 0.176 atm

Explanation: Solution attached:

Use Boyle's Law to find the new pressure of the gas.

P1V1 = P2V2

Derive for P2

P2 = P1V1 / V2

= 5.5 atm ( 4.8 L ) / 150 L

= 0.176 atm

4 0

2 years ago

To what volume should 25ml of 15m nitric acid be diluted to prepare a 3m solution

Maurinko [17]

We can use the dilution formula to find the volume of the diluted solution to be prepared

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting the values in the equation

15 M x 25 mL = 3 M x v2

v2 = 125 mL

The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution

6 0

3 years ago

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

2 years ago

A block of ice has edge lengths of 8.00 cm each and a mass of 476

valkas [14]

Density is calculated as mass divided by volume. If we are given an ice cube of side length 8.00 cm, then the volume of the cube is equivalent to (8.00 cm)^3 = 512 cm^3. Since we have a given mass of 476 g, we can divide:
476 g / 512 cm^3 = 0.930 g/cm^3
So the density of the sample of ice is 0.930 g/cm^3.

3 0

2 years ago