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.
I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge.
In the reaction as follows: NH2- + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowrybase.
BRØNSTED-LOWRY BASE:
According to Bronsted-Lowry definition of a base and acid, a base is substance that accepts an hydrogen ion or proton (H+) while an acid is a substance that donates a proton.
According to this reaction given as follows: NH2 + CH3OH → NH3+ CH3O-
NH2- is a reactant that accepts a hydrogen ion (H+) to become NH3+
NH3+CH3OH is a reactant that donates hydrogen ion (H+)
Since NH2- accepts a proton, this means that in the reaction as follows: NH2 + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowrybase.
The heat that is required to raise the temperature of an object is calculated through the equation, heat = mass x specific heat x (T2 - T1) Specific heat is therefore calculated through the equation below, specific heat = heat / (mass x (T2 - T1)) Substituting, specific heat = 645 J / ((28.4 g)(15.5 - - 11.6)) The value of specific heat from above equation is 0.838 J/g°C.
