Hemoglobin contains 0.25% iron/weight the molecular mass of hemoglobin is 89600. Calculate the no. of helium atoms present per molecule of hemoglobin.
Answers
Correct Question :
Haemoglobin contains 0.25% iron/weight the molecular mass of hemoglobin is 89600 calculate the no. of Iron (Fe) atoms present per molecule of hemoglobin?
Solution :
Molecular mass of Haemoglobin = 89600
Percentage of Iron = 0.25% / weight. This means the total mass of Iron out of the total molecular mass of Haemoglobin is 0.25% of 89600.
Given mass of Iron,
⇒ 0.25% × 89600
⇒ 25/10000 × 89600
⇒ 1/400 × 89600
⇒ 224 gm of Iron
Moles of Iron,
⇒ Given Mass / Atomic mass
⇒ 224 gm / 56 gm
⇒ 4 moles
Now here, 1 mole of Haemoglobin contains 4 moles of Iron (Fe). Then 1 molecule of Haemoglobin will have 4 atoms of Fe.
Hence, no. of Iron atoms present per molecule of hemoglobin is 4.
Answer:
nswer :
Iron. The blue colour of aqueous copper sulphate solution can be changed to place green by immersing an iron rod in it.
Explanation :
Out of the given options, all except silver can displace copper from its solution. But, we are also asked that the solution should be changed to a colour of pale-green. Hence, the answer is iron beacuse the colour of iron sulphate solution is pale-green.
When an iron rod is immersed in the copper sulphate solution it displaces copper from its solution and forms iron sulphate solution which gives a pale green colour to the resultant solution.
Reaction :
\sf{CuSO_4(aq)+Fe(s)\longrightarrow{}\underset{\textsf{pale green}}{FeSO_4(aq)}+Cu(s)}CuSO4(aq)+Fe(s)⟶pale greenFeSO4(aq)+Cu(s)
_____________
Additional Information :
Displacement reaction :
A type of chemical reaction in which an atom or group of atoms displace another atom or group of atoms of a compound. Displacement reactions are also called single-displacement reactions.
Examples :
\sf{CuCl_2(aq)+Zn(s)\longrightarrow{}ZnCl_2(aq)+Cu(s)}CuCl2(aq)+Zn(s)⟶ZnCl2(aq)+Cu(s)
\sf{CuSO_4(aq)+Zn(s)\longrightarrow{}ZnSO_4(aq)+Cu(s)}CuSO4(aq)+Zn(s)⟶ZnSO4(aq)+Cu(s)
\sf{CuCl_2(aq)+Pb(s)\longrightarrow{}PbCl_2(aq)+Cu(s)}CuCl2(aq)+Pb(s)⟶PbCl2(aq)+Cu(s)