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3 years ago

Which of the following statements is not part of Dalton's atomic theory? ( points)

1 answer:

4 0

The answer is "different atoms come together to form an element"- it is wrong because elements are made up of the same atoms.

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If a gas in the above system has a volume of V at a pressure of P, what is the volume of the same gas at a pressure of 4p

gladu [14]

Answer: 0.25V

Explanation:

P1V1 = P2V2

PV = (4P)V2

V2 = 0.25V

8 0

2 years ago

or the solution containing Zn2+(aq), n is 2, so the MM/n ratio will be the molar mass of Zn divided by 2: MM/n = (65.39 g/mol) /

yKpoI14uk [10]

Answer:

MM/n will be less than the theoretical value

Explanation:

Provided that

The plating efficiency is proportional to the element weight

i.e.

$\frac{Zn^2}{(aq)}$

That represents $\frac{MM}{n\ value}$

So,

MM by N value is

= 65.39 g /mol ÷ 2

= 32.70 g/mol

Also, the experimental plating value is 0.55 that represents that it is much lesser than the normal value

So it would be concluded that the MM by N would be lower than the theoretical value

4 0

2 years ago

Calculate the final concentration of each of the following:

kkurt [141]

Answer:

1. 2 M

2. 2 M

Explanation:

1. Determination of the final concentration.

Initial Volume (V₁) = 2 L

Initial concentration (C₁) = 6 M

Final volume (V₂) = 6 L

Final concentration (C₂) =?

The final concentration can be obtained as follow:

C₁V₁ = C₂V₂

6 × 2 = C₂ × 6

12 = C₂ × 6

Divide both side by 6

C₂ = 12 / 6

C₂ = 2 M

Therefore, the final concentration of the solution is 2 M

2. Determination of the final concentration.

Initial Volume (V₁) = 0.5 L

Initial concentration (C₁) = 12 M

Final volume (V₂) = 3 L

Final concentration (C₂) =?

The final concentration can be obtained as follow:

C₁V₁ = C₂V₂

12 × 0.5 = C₂ × 3

6 = C₂ × 3

Divide both side by 3

C₂ = 6 / 3

C₂ = 2 M

Therefore, the final concentration of the solution is 2 M

7 0

2 years ago

In Thomson’s model of the atom , what parts of the atom were in motion

trasher [3.6K]

Answer:

The electrons where in motion

4 0

3 years ago

The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nu

andreyandreev [35.5K]

Answer:

The binding energy per nucleon = 1.368*10^-12 (option D)

Explanation:

<u>Step 1:</u> Data given

The mass of a proton is 1.00728 amu

The mass of a neutron is 1.00867 amu

The mass of a cobalt-60 nucleus is59.9338 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt60 has 27 protons and 33 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 33 neutrons = 33*1.00867 u = 33.28611 u

Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u

Mass of a cobalt60 nucleus = 59.9338 amu

Mass defect = Δm = 0.54887 u

ΔE =c²*Δm

ΔE = (3.00 *10^8 m/s)² *(0.54887 amu))*(1.00 g/ 6.02 *10^23 amu)*(1kg/1000g)

Step 3: Calculate binding energy per nucleon

ΔE = 8.21 * 10^-11 J

8.21* 10^-11 J / 59.9338 = 1.368 *10^-12

The binding energy per nucleon = 1.368*10^-12 (option D)

3 0

3 years ago