What two forces act in the nucleus to create a 'nuclear tug of war'?

Answers

Answer 1

Answer: In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons. This nuclear force is a short range force and is so strong that it negates the effect of the force of repulsion between the protons and helps bind the nucleus together. Hope this helped :D

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NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)Using the above formula, if I used 0.100L of 0.83 M HC2H3O2 (aq) (0.83 mol per 1 L of solution), how much carbon dioxide would be produced?

Answers

Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams

Final answer:

By utilizing stoichiometry, we can determine that the given volume and molarity of acetic acid (HC2H3O2) would produce approximately 3.65 grams of carbon dioxide (CO2). The calculation involves determining the moles of HC2H3O2 used, which equals the moles of CO2 produced, and converting that to grams using the molar mass of CO2.

Explanation:

The amount of carbon dioxide produced can be determined through stoichiometry, using the provided balanced chemical equation and molarity (M) of the acetic acid HC2H3O2. According to the balanced chemical equation, one mole of NaHCO3 reacts with one mole of HC2H3O2 to produce one mole of CO2. In other words, the moles of HC2H3O2 used equals the moles of CO2 produced.

First, calculate the moles of HC2H3O2 by multiplying the given volume (0.100 L) by its molarity (0.83 mol/L): 0.100L x 0.83 mol/L = 0.083 mol.

So, according to the stoichiometric ratio, 0.083 mol of HC2H3O2 would produce 0.083 mol of CO2. If you want your answer in grams, note that the molar mass of CO2 is approximately 44.01 g/mol. Multiply the moles of CO2 by its molar mass: 0.083 mol x 44.01 g/mol = 3.65g.

So, the amount of CO2 produced would be 3.65 grams.

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as a result of the gold foil experiment it was concluded that an atom (1) contains protons, neutrons, and electrons(2)contains a small, dense nucleus(3)has positrons and orbitals (4)is a hard and invisible sphere

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It showed that the atom contains a small dense nucleus. The reason why this can be concluded is that the experiment showed that alpha particles can for the most part pass right through the foil. This showed that the nucleus of the atom is very small and takes up a very small portion of the size of an atom since the nucleus is what would reflect the alpha particle. If atoms were not mostly empty space, the alpha particles would have been bounced back and the opposite of what was observed would have happened (all or at least the majority would be bounced back instead the majority getting through with only a small portion being reflected back).

I hope this helps. Let me know in the comments if anything is unclear.

Final answer:

The gold foil experiment, conducted by Ernest Rutherford, discovered the existence of a small, dense nucleus within an atom, making answer (2) the correct one.

Explanation:

The gold foil experiment was conducted by Ernest Rutherford and led to the discovery of a small, dense nucleus within the atom. This means that the correct answer to your question is (2) contains a small, dense nucleus. Prior to this experiment, the widely accepted model was the 'plum pudding' model, which had negatively charged electrons embedded within a positively charged sphere. The gold foil experiment upended this model as it was discovered that most alpha particles passed through the gold foil undeterred, but a small fraction was deflected at large angles, suggesting that a tiny but dense packet of positive charge — identified as the nucleus — existed in the center of the atom. This discovery led to the formation of the Rutherford atomic model.

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Rank the following structures in order of decreasing electrophilic strength.

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Electrophiles are positively charged or neutral species having empty orbitals that are attracted to an electron rich center. Examples of these are the Bronsted acid. Bronsted acid accepts an electron pair for hydrogen.

the relative atomic mass of boron 10.2.if the element occurs in two isotopic forms,boron-10 and boron-11 ,what are their geometrical abundances?

Answers

The correct answer for the question that is being presented above is this one:

Boron consists of two isotopes, boron-10 and boron-11.
Given that its relative atomic mass is 10.2

Find the abundance of each isotope.

Let y/100 = the abundance of copper-10
and (100 - y)/100 = the abundance of copper-11

10.2 = (y/100 x 10) + [(100 - y)/100 x 11]
10.2 = 10y/100 + 1100/100 - 11y/100
1020 = 10y + 1100 - 11y
-80 = -y
y = 80

Abundance of boron-10 = 10/100 = 10%

Abundance of boron-11 = 100 - 10 = 90%

In chemical reactions, there is a conservation of(1) energy, volume and mass(2) energy, volume and charge
(3) energy, charge and mass
(4) energy, charge and volume

Answers

Answer: option (3) energy, charge and mass.

Explanation:

1) The conservation of energy is a universal principle: energy is neither created nor destroyed. This is the first law of thermodynamic.

2) Mass conservation is another universal principle: mass is neither created nor destroyed. This principle is the base of the stochiometry: thas mass of the reactans equal the mass of the products. Another consequence, since the atoms are not transformed in the chemical reactions, is that the number of each kind of atoms in the reactants equal the number of the same kind of atoms in the products.

3) The third principle is the conservation of charge. Also, charge is neither created nor destroyed. The electrons gained/lost by one species are lost/gained by other species. So, when one atom or molecule is oxidized other is reduced.

(3) energy, charge, and mass

What are two elements that form a large number of bonds and that both have similar properties to carbon?

Answers

Silicon and Germanium. You can conclude it from its position in the periodic table. Silicon and Germanium are in the same group of the Carbon. That implies that all three have the same number of electrons in the outer orbital, which is responsible for the bonds.

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