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3 years ago

Match the scientist to his contribution to the atomic theory. Match Term Definition J.J. Thomson A) Positively charged nucleus

Chemistry

2 answers:

marissa [1.9K]3 years ago

6 0

Answer is:

J.J. Thomson- B) Plum Pudding Model.

J. J. Thomson discovered the electron in 1897.

His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.

John Dalton - C) Atoms indivisible.

Dalton stated that atom is matter that can not be divided

Ernest Rutherford - A) Positively charged nucleus.

According to Rutherford model of the atom:

1) Atoms have their charge concentrated in a very small nucleus.

2) Major space in an atom is empty.

3) Atoms nucleus is surrounded by negatively charged particles called electrons.

4) An atom is electrically neutral.

James Chadwick - D) Discovery of neutrons.

James Chadwick was a British physicist who was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron in 1932.

pantera1 [17]3 years ago

5 0

J.J. Thomson :Plum Pudding Model

James Chadwick: Discovery of neutrons

John Dalton:Atoms indivisible

Ernest Rutherford:Positively charged nucleus

I hope this helps and have a great day!

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determine the concentration of 35.0mL of an acid solution that requires 25.1mL of a 0.50M NaOH to neutralize it

Degger [83]

Hello!

The general chemical equation for the neutralization of an acid HA with NaOH is the following:

HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

For determining the concentration of the acid solution, we can use the equation shown below:

$moles HA=moles NaOH \\ \\ cHA*VHA=cNaOH*VNaOH \\ \\ cHA= \frac{cNaOH*vNaOH}{VHA}= \frac{0,5M*25,1 mL}{35 mL}=0,36 M$

So, the concentration of the Acid is 0,36 M

Have a nice day!

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4 years ago

How many grams of iron metal do you expect to be produced when 325 grams of an 87.5 percent by mass iron (II) nitrate solution r

Debora [2.8K]

Answer: 88.2 g

Solution:

1) Chemical equation:

<span>2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)3 (aq)

2) Theoretical molar ratios

2 mol Al : 3 mol Fe(NO3)2 : 3 mol Fe : 2 mol Al(NO3)3

3) Starting mass of pure iron nitrate

% = (mass of iron nitrate / mass of solution) * 100 = 87.5

=> mass of iron nitrate = 87.5 * mass of solution / 100

mass of solution = 325 g

=> mass of iron nitrate = 87.5 * 325 g / 100 = 284.375 g

4) moles of iron nitrate

moles = mass in grams / molar mass

molar mass of Fe(NO3)2 = 179.85 g/mol

moles = 284.375 g/ 179.85 g/mol = 1.58 moles Fe(NO3)2

5) proportion:

             x                                 3 mol Fe
--------------------------- =     ----------------------
1.58 mol Fe(NO3)2          3 mol Fe(NO3)2

Clear x:

x = 1.58 mol Fe

6) Convert 1.58 mol Fe into grams

mass = number of moles * atomic mass

atomic mass of iron = 55.845 g / mol

mass = 1.58 moles * 55.845 g/mol = 88.24 g

Rounded to 3 significant figures: 88.2 grams of Fe.
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Answer:1.

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