Chemistry, 15.07.2020 01:01 cornpops4037

The average human body contains 5.00 L of blood with a Fe2+ concentration of 1.10×10−5 M . If a person ingests 9.00 mL of 21.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

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Answer from: allieballey0727

The percentage is % $Fe^{2+$ = 57.3 %

Explanation:

From the question we are told that

The volume of blood in the human body is $V = 5.0 0 \ L$

The concentration of $Fe^{2+$ is $C_{F} = 1.10 *10^{-5} \ M$

The volume of NaCN ingested is $V_N = 9.00 \ mL = 9.00 *10^{-3} \ L$

The concentration of NaCN ingested is $C_N = 21.0 \ mM = 21.0 *10^{-3} \ M$

The number of moles of $Fe^{2+$ in the blood is

N_F = C_F * V

substituting values

$N_F = 1.10 *10^{-5} * 5$

$N_F = 5.5*10^{-5} \ mols$

The number of moles of $CN^{-}$ ingested is mathematically evaluated as

N_C = C_N * V_N

substituting values

$N_C = 21*10^{-3} * 9 *10^{-3}$

$N_C = 1.89 *10^{-4} \ mols$

The balanced chemical equation for the reaction between $Fe^{2+$ and $CN^{-}$ is represented as

$Fe^{2+} + 6 CN^{-} \to [Fe(CN)_6]^{2-}$

From this reaction we see that

1 mole of $Fe^{2+$ will react with 6 moles of $CN^{-}$

=> x moles of $Fe^{2+$ will react with $1.89 *10^{-4} \ moles$ of $CN^{-}$

Thus

$x = \frac{1.89 *10^{-4} * 1}{6}$

$x = 3.15 *10^{-5}$

Hence the percentage of $Fe^{2+$ that reacted is mathematically evaluated as

% $Fe^{2+$ $= \frac{3.15 *10^{-5}}{5.5*10^{-5}} * 100$

% $Fe^{2+$ = 57.3 %

Answer from: Quest

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

Answer from: Quest

Answer246 k;

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