The role of the gluons is to bind together the quarks that make up protons and neutrons. That is option D.
<h3>What are gluons?</h3>
Gluons are those particles found within an atom that are responsible for binding protons and neutrons together inside the nucleus of an atom.
The gluons are capable of binding the protons and neutrons together by holding together the quarks that makes up protons and neutrons.
These gluons are known to carry a color-anticolor charge which makes up the 9 types of gluons which include:
An atom is the smallest indivisible part of an element that can take part in a chemical reaction.
The proton is the positive particle of an atom which is found within the nucleus.
The neutron is the particle that is found in the nucleus of an atom which doesn't have any charge.
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6 because the number of valence electrons is always the ones-place of any two-digit group number. So an element in group 18 would have 8 valence electrons
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Answer:
A conversion factor is a variable that transforms the dimension of one variable into the terms of another variable.
Explanation:
Let be 
and 
two variables with diferent dimensions (units), 
is a conversion if transforms the dimension of one variable into the terms of the other variable. That is:
(1)
Where:
- Input, [input unit]
- Output, [output unit]
- Conversion factor, [output unit] per [input unit]
Stuff about the earth and tectonic plates, and about space and stuff.
1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.
2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂<span>(g).
m(HgO) = 4.37 g.
n</span>(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = <span>0.02 mol; amount of substance.
m</span>(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
<span>yield = 61.77%.
3. Answer is: 68.16 </span><span>grams of the excess reactant (oxygen) remain.
</span>Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃<span>(g).
m(Fe) = 27.3 g.
n</span>(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.
4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).
5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
<span>Empirical
formula gives the proportions of the elements present in a compound.
</span>Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol.
6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (<span>number of particles (ions,</span> atoms<span> or </span>molecules), that are contained in <span>one </span>mole of substance<span>).
</span>
7. Answer is: iron (Fe) <span>is the limiting reactant.
</span>Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.
8. Answer is: there
are 0.435
moles of C₆H₁₄.<span>
N(C₆H₁₄) = 2.62·10²³; number of molecules.</span><span>
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.</span><span>
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.</span><span>
n(C₆H₁₄) = 0.435 mol; amount of substance of </span>C₆H₁₄.<span>
Na - Avogadro constant or Avogadro number.
</span>
9. Answer is: 3.675 <span>moles of carbon(II) oxide are required to completely react.
</span>Balanced chemical reaction: Fe₂O₃<span>(s) + 3CO(g) ⟶ 2Fe(s) + 3CO</span>₂<span>(g).
n(</span>Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.
10. Answer is: there are 2.158 moles of barium atoms.<span>
N(Ba</span><span>) = 2.62·10²³; number of
atoms of barium.
n</span>(Ba) = N(Ba)<span> ÷ Na.
n</span>(Ba) = 1.3·10²⁴<span> ÷
6.022·10²³ 1/mol.
n</span>(Ba)<span> = 2.158 mol; amount of
substance of barium</span>.<span>
Na - Avogadro constant or Avogadro number.
11. Answer is: </span>6.26·10²³ <span>carbon atoms are present.
</span>n(C₂H₆O) = 0.52 mol; amount of substance.<span>
N</span>(C₂H₆O) =
n(C₂H₆O) · Na.<span>
N</span>(C₂H₆O) =
0.52 mol · 6.022·10²³ 1/mol.<span>
N</span>(C₂H₆O) =
3.13·10²³.<span>
In one molecule of </span>C₂H₆O there are two atoms of carbon:<span>
N(C</span>) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.
<span>
12. Answer is: </span><span>the empirical formula is C</span>₂H₄O.<span>
</span><span>If we use 100 grams of compound:
</span>1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.
