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Butoxors [25]
3 years ago
6

A chemist adds of a M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide th

e chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

0.42 mg

Explanation:

It seems the question is incomplete, however I'll use values that have been found in a web search:

"A chemist adds 55.0 mL of a 1.7x10⁻⁵ M mercury(II) iodide solution to a reaction flask. Calculate the mass in milligrams of mercury(II) iodide the chemist has added to the flask. Round your answer to 2 significant digits."

<u>Keep in mind that while the process of solving the problem remains the same, if the values in your question are different, your answer will differ as well</u>.

First we <u>calculate the moles of mercury (II) iodide</u> (HgI₂):

  • 1.7x10⁻⁵M * 55.0 mL = 9.35x10⁻⁴ mmol HgI₂

Then we<u> convert mmol to mg, using the molar mass</u> (454.4 g/mol):

  • 9.35x10⁻⁴ mmol HgI₂* 454.4 mg/mmol = 0.42 mg
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Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96
cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0
3 years ago
Write Fe salts (divalent Fe)?​<br><br>Pleasee help.... I will mark the answer with brainlist.
kakasveta [241]

Answer:

Graphitic carbon nitride (g-C3N4) is a rising two-dimensional material possessing intrinsic semiconducting property with unique geometric configuration featuring superimposed heterocyclic sp2 carbon and nitrogen network, nonplanar layer chain structure, and alternating buckling. The inherent porous structure of heptazine-based g-C3N4 features electron-rich sp2 nitrogen, which can be exploited as a stable transition metal coordination site. Multiple metal-functionalized g-C3N4 systems have been reported for versatile applications, but local coordination as well as its electronic structure variation upon incoming metal species is not well understood. Here we present detailed bond coordination of divalent iron (Fe2+) through micropore sites of graphitic carbon nitride and provide both experimental and computational evidence supporting the aforementioned proposition.

5 0
3 years ago
Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the foll
Nana76 [90]

Answer: C)  0.637 V

Explanation:

The balanced redox reaction is:

Zn(s)+Pb^{2+}(aq)\rightarrow Zn^{2+}(aq)+Pb(s)

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

Zn^{2+}(aq)+2e^-\rightarrow Zn(s)= -0.763

Pb^{2+}(aq)+2e^-\rightarrow Pb(s)= -0.126

E^0_{[Zn^{2+}/Zn]}=-0.763V

E^0_{[Pb^{2+}/Pb]}=-0.126V

E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}

E^0=-0.126-(-0.763V)=0.637V

The standard emf of a cell is 0.637 V

8 0
3 years ago
What will litmus paper turn when it reacts with sodium nitrate?​
castortr0y [4]
A solution of sodium nitrate will get make litmus paper wet. A solution of NaNO3 is neutral. Therefore, there is no change in color.
6 0
3 years ago
The domain Archaea are unicellular prokaryotes and can be autotrophs or heterotrophs
nexus9112 [7]

Answer:

I think its true I dont really know

Explanation:

true

7 0
3 years ago
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