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Veronika [31]
3 years ago
10

Complete the following table for the three key subatomic particles.

Chemistry
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer:

This question is incomplete without the table

Explanation:

The missing table is attached to this answer.

There are three subatomic particles found in an atom; protons, neutrons and electrons. The properties of the subatomic particles will be classified under the following subheadings (which are the empty columns/boxes in the attachment)

SYMBOL

The symbol of proton is "p⁺", the symbol of neutron is "n⁰" and the symbol of electron is "e⁻".

CHARGE

Looking at the symbol of the subatomic particles, one can guess the charge of each of the particles from the superscript.

Protons are positively charged, electrons are negatively charged while neutrons have no charge/electrically neutral.

The relative charge of proton is +1 while it's absolute charge is +1.60218 × 10⁻¹⁹

The charge of a proton was first determined by Ernest Rutherford using the gold-foil experiment

The relative charge of electron is -1 while it's absolute charge is -1.60218 × 10⁻¹⁹

The charge of an electron was first determined by R. Milikan using the oil-drop experiment

The relative charge and absolute charge of neutron is 0

The charge of a neutron was first determined by (or credited to) James Chadwick.

MASS

The relative mass (amu) of proton is 1.00727 while that of neutron is 1.00866. The relative mass (amu) of an electron is 0.00054858 while it's absolute mass (g) is 9.10939 × 10⁻²⁴.

The mass of an electron was first measured by J. J. Thomson.

The mass of a proton was first measured by (or credited to) Ernest Rutherford.

The mass of a neutron was first measured by James Chadwick.

LOCATION

The protons and neutrons <u>are located inside the nucleus</u> which is found in the centre of an atom while the electron(s)<u> is/are found outside the nucleus but within the atom</u>.

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The following half-reaction can be balanced in acidic solution:
den301095 [7]

Answer:

None of them, answer seems to be 24 unless I messed up

Explanation:

How many electrons appear in the balanced equation

What the heck do they mean, do they mean transferred?

N2O5 --> NH4+

Left Side

N 5+  Electrons 2 in central shell, 5 in outer

O 2- Electrons 2 in central shell, 6 in outer

Right Side

H  1+ Electrons 1 in central shell

N 3- Electrons 2 in central shell, 5 in outer

Hmm, ok so need to balance the half equation first, this is a redox reaction.

Hmm, so it's going to be something like

H2 + N2O5 → NH + O3  

Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

H2 + N2O5 → NH + O3  

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

H02 + N+52O-25 → N-1H+1 + O03  

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

When one member of the redox couple is oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it is best to replace it with a water molecule.

O:3H+12O-2 → O03 + 6e-(O)

H02 → H+12O-2 + 2e-(H)

R:N+52O-25 + 12e- → 2N-1H+1(N)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1  

c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O  

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+| *3

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O| *2

O:9H+12O-2 + 3H02 + 3H2O → 3O03 + 3H+12O-2 + 24e- + 24H+  

R:2N+52O-25 + 24e- + 24H+ → 4N-1H+1 + 10H2O  

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

9H+12O-2 + 2N+52O-25 + 3H02 + 24e- + 3H2O + 24H+ → 3O03 + 4N-1H+1 + 13H2O + 24e- + 24H+

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2N+52O-25 + 3H02 → 3O03 + 4N-1H+1 + H2O

7 0
3 years ago
What are some important things in this paragraph ???? BRAINLIST WILL BE AWARDED!
mezya [45]

Answer:

Suppose you were standing in the dark room with a bare light bulb behind you. You hold a ball in front of you, and you can see all of the lit half of the ball, which looks like a circle.

you rotate 90° to your left, and see the left side of the circle lit while the right side is dark. Half the ball is still lit up, but you can see only part of the lit area. As you continue to rotate, you see a different amount of the ball. In fact, you would see the shape change from a full circle to a crescent shape to a backwards crescent shape and then back to a full circle.

Explanation:

I think this is it tell me if i'm wrong

6 0
3 years ago
An ideal gas is kept in a 10​-liter ​[L] container at a pressure of 2.5 atmospheres​ [atm] and a temperature of 310 kelvin​ [K].
Hatshy [7]

Answer:

5 L.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 10 L

Initial pressure (P1) = 2.5 atm

Final pressure (P2) = 5 atm

Final volume (V2) =.?

Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:

P1V1 = P2V2

2.5 × 10 = 5 × V2

25 = 5 × V2

Divide both side by 5

V2 =25/5

V2 = 5 L

Thus, the new volume of the gas is 5 L

8 0
2 years ago
What was the hiker's average velocity during part C of the hike?
umka2103 [35]

Answer:

0.13 km/min west

Explanation:

I just did the test.

6 0
2 years ago
I need to make up an experiment !! Please help
katen-ka-za [31]

Question: Baking a Cake Without Flour.

Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.

Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.  

Results: My control cake, which I cooked for 25 minutes, measured 4 cm high.  Eight out of ten tasters that I picked at random from the class found it to be an acceptable dessert. After 25 minutes of baking, my experimental cake was 1.5 cm high and all ten tasters refused to eat it because it was burnt to a crisp.

What did I learn?/Conclusion: Since the experimental cake burned, my results did not support my hypothesis.  I think that the cake burned because it had less mass, but cooked for the same amount of time.  I propose that the baking time be shortened in subsequent trials.

-

I hope this helped :))

7 0
3 years ago
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