Answer:
The proton remains the same.
Explanation:
Oxidation is simply defined as the loss of electron(s) during a chemical reaction either by an atom, molecule or ion.
Oxidation is strictly on the transfer of electron(s) and not proton.
A metal that undergoes oxidation still has its protons intact otherwise it will not be called the ion of the metal since atomic number is called the proton number.
Sodium (Na) undergoes oxidation as follow:
Na —> Na+ + e-
Na is called sodium metal.
Na+ is called sodium ion.
Na has 11 electrons and 11 protons
Na+ has 10 electrons and 11 protons
From the above illustration, we can see that the protons of Na and Na+ are the same why their electrons differ because Na+ indicates that 1 electron has been loss or transferred.
Explanation:
a.
→ ?
![Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%28aq%29%20%2B%20Na_2SO_4%28aq%29%5Crightarrow%20PbSO_4%28s%29%2B2NaNO_3%28aq%29)
![Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%28aq%29%5Crightarrow%20Pb%5E%7B2%2B%7D%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29)
![Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)](https://tex.z-dn.net/?f=Na_2SO_4%28aq%29%5Crightarrow%202Na%5E%2B%2BSO_4%5E%7B2-%7D%28aq%29)
![Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%28aq%29%2B2NO_3%5E%7B-%7D%28aq%29%2B2Na%5E%2B%2BSO_4%5E%7B2-%7D%28aq%29%5Crightarrow%20PbSO_4%28s%29%2B2Na%5E%2B%2B2NO_3%5E%7B-%7D%28aq%29)
Removing common ions from both sides, we get the net ionic equation:
![Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)](https://tex.z-dn.net/?f=Pb%5E%7B2%2B%7D%28aq%29%2BSO_4%5E%7B2-%7D%28aq%29%5Crightarrow%20PbSO_4%28s%29)
b.
→
![NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)](https://tex.z-dn.net/?f=NiCl_2%28aq%29%20%2B%20NH4NO_3%28aq%29%20%5Crightarrow%20Ni%28NO_3%29_2%2BNH_4Cl%28aq%29)
No precipitation is occuring.
c.
→
![FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)](https://tex.z-dn.net/?f=FeCl_2%28aq%29%20%2B%20Na_2S%28aq%29%5Crightarrow%20FeS%28s%29%2B2NaCl%28aq%29)
![FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)](https://tex.z-dn.net/?f=FeCl_2%28aq%29%5Crightarrow%20Fe%5E%7B2%2B%7D%28aq%29%2B2Cl%5E%7B-%7D%28aq%29)
![Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)](https://tex.z-dn.net/?f=Na_2S%28aq%29%5Crightarrow%202Na%5E%2B%2BS%7B2-%7D%28aq%29)
![Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%28aq%29%2B2Cl%5E%7B-%7D%28aq%29%2B2Na%5E%2B%2BS%5E%7B2-%7D%28aq%29%5Crightarrow%20FeS%28s%29%2B2Na%5E%2B%2B2Cl%5E%7B-%7D%28aq%29)
Removing common ions from both sides, we get the net ionic equation:
![Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%28aq%29%2BS%5E%7B2-%7D%28aq%29%5Crightarrow%20FeS%28s%29)
d.
→
![MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2](https://tex.z-dn.net/?f=MgSO4%28aq%29%20%2B%20BaCl2%28aq%29%5Crightarrow%20BaSO_4%28s%29%2BMgCl_2)
![MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)](https://tex.z-dn.net/?f=MgSO_4%28aq%29%5Crightarrow%20Mg%5E%7B2%2B%7D%28aq%29%2BSO_4%5E%7B2-%7D%28aq%29)
![BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)](https://tex.z-dn.net/?f=BaCl_2%28aq%29%5Crightarrow%20Ba%5E%7B2%2B%7D%2B2Cl%5E%7B-%7D%28aq%29)
![Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)](https://tex.z-dn.net/?f=Mg%5E%7B2%2B%7D%28aq%29%2BSO_4%5E%7B2-%7D%28aq%29%2BBa%5E%7B2%2B%7D%2B2Cl%5E%7B-%7D%28aq%29%28aq%29%5Crightarrow%20BaSO_4%28s%29%2BMg%5E%7B2%2B%7D%28aq%29%2B2Cl%5E%7B-%7D%28aq%29)
Removing common ions from both sides, we get the net ionic equation:
![Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)](https://tex.z-dn.net/?f=Ba%5E%7B2%2B%7D%28aq%29%2BSO_4%5E%7B2-%7D%28aq%29%5Crightarrow%20BaSO_4%28s%29)
Answer:
The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol
Explanation:
Fractional distillation is a separation process based on difference in boiling point of two compounds.
1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.
1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.
Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.
So, boiling point of 1-butanol is much higher than 1-chlorobutane.
Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.
So, option (D) is correct.
Answer:
The mass of copper(II) sulfide formed is:
= 81.24 g
Explanation:
The Balanced chemical equation for this reaction is :
![Cu(s) + S\rightarrow CuS](https://tex.z-dn.net/?f=Cu%28s%29%20%2B%20S%5Crightarrow%20CuS)
given mass= 54 g
Molar mass of Cu = 63.55 g/mol
![Moles = \frac{given\ mass}{Molar\ mass}](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bgiven%5C%20mass%7D%7BMolar%5C%20mass%7D)
Moles of Cu = 0.8497 mol
Given mass = 42 g
Molar mass of S = 32.06 g/mol
![Moles = \frac{given\ mass}{Molar\ mass}](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bgiven%5C%20mass%7D%7BMolar%5C%20mass%7D)
Moles of S = 1.31 mol
Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n
<u><em>First find the limiting reagent :</em></u>
![Cu + S\rightarrow CuS](https://tex.z-dn.net/?f=Cu%20%2B%20S%5Crightarrow%20CuS)
1 mol of Cu require = 1 mol of S
0.8497 mol of Cu should require = 1 x 0.8497 mol
= 0.8497 mol of S
S present in the reaction Medium = 1.31 mol
S Required = 0.8497 mol
S is present in excess and <u>Cu is limiting reagent</u>
<u>All Cu is consumed in the reaction</u>
Amount Cu will decide the amount of CuS formed
![Cu + S\rightarrow CuS](https://tex.z-dn.net/?f=Cu%20%2B%20S%5Crightarrow%20CuS)
1 mole of Cu gives = 1 mole of Copper sulfide
0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide
= 0.8497
Molar mass of CuS = 95.611 g/mol
![Moles = \frac{given\ mass}{Molar\ mass}](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bgiven%5C%20mass%7D%7BMolar%5C%20mass%7D)
![0.8497 = \frac{given\ mass}{95.611}](https://tex.z-dn.net/?f=0.8497%20%3D%20%5Cfrac%7Bgiven%5C%20mass%7D%7B95.611%7D)
Mass of CuS = 0.8497 x 95.611
= 81.24 g
Answer:
The correct answer will be "4.60 g".
Explanation:
The given values are:
Volume of Butane = 7.96 mL
Density = 0.579 g/mL
As we know,
⇒ ![Mass \ of \ Butane = Density\times Volume](https://tex.z-dn.net/?f=Mass%20%5C%20of%20%5C%20Butane%20%3D%20Density%5Ctimes%20Volume)
On putting the estimated values, we get
⇒ ![=0.579\times 7.96](https://tex.z-dn.net/?f=%3D0.579%5Ctimes%207.96)
⇒ ![=4.60 \ g](https://tex.z-dn.net/?f=%3D4.60%20%5C%20g)