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AlladinOne [14]
3 years ago
5

Differentiate between a single atom of gold and a single atom of lead. (No Multiple Choice)

Chemistry
1 answer:
tatyana61 [14]3 years ago
7 0
Lead has a different chemistry than most biological metals, but it can mimic some of them, such as calcium, iron, and zinc, because it tends to form 2+ ion like other metals, and these can be formed inside the body. Due to it different chemistry, lead prevents normal functioning of the enzymes it binds to, inhibiting them and poisoning the body.


Gold rarely forms ions, unless under extreme conditions that are not present in the human body. Gold ions, when ingested, are actually very toxic, so gold poisoning is possible. But because gold are so rare, gold poisoning is also a rare phenomenon.
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3 years ago
During early research in nuclear energy, why were scientists surprised that the products from splitting a uranium nucleus had le
creativ13 [48]
Scientists expected that the law of conservation of mass would apply to nuclear fission in terms of the masses of the subatomic particles. In reality, the mass of an atom is not equal to the sum of the masses of the subatomic particles that make it up. This is because of the energy that binds the subatomic particles. This energy has mass and when the bond is broken, the mass of the energy of the bonds is lost resulting to what we now cal, a mass defect.
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3 years ago
Read 2 more answers
If a 185-lb patient is prescribed 145mg. What dosage is the patient receiving in mg/kg of his body weight?
Anestetic [448]

Answer:

1.728 mg/kg

Explanation:

First have to convert the patient's weight to kg

1 lb = 0.454 kg

185 lb in kg = 83.9 kg

Then we have to divide the dose per the weight of the patient

145 mg for 83.9 kg =  145/83.9

= 1.728 mg/kg

7 0
3 years ago
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4 0
2 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
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