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2 years ago

Which of the following would be found in both eukaryotic AND prokaryotic cells?

Chemistry

2 answers:

beks73 [17]2 years ago

7 0

Nucleus ,endoplasmic reticulum

ladessa [460]2 years ago

7 0

Answer:

Explanation:

the Nucleus and Endoplasmic Reticulum are both found in both eukaryotic and prokaryotic cells, but the necleus is the more common answer.

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what is the correct conversion factor by which to multiply to convert moles of nitrogen to moles of ammonium nitrate (NH4NO3)

Ostrovityanka [42]

Answer:

0.5

Explanation:

1 mole of ammonium nitrate contains 2 moles of nirogen

1 mole of nitrogen converts to 0.5 moles of ammonium nitrate

the conversation factor is 0.5

6 0

3 years ago

Read 2 more answers

How many grams is 10 moles of H2O?

Elan Coil [88]

Answer:

So, 10 mole of water will weigh (18x10) = 180g.

5 0

2 years ago

Chromium-51 is a radioactive isotope used to label red blood cells to identify if they remain intact in the body. It has a half-

agasfer [191]

$\left[\begin{array}{cccc}0 \ days&27,7days&55,4days&\textbf{83,1days}\\&&&\\0,25mg&0,125mg&0,0625mg&\textbf{0,03125mg}\end{array}\right]$

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3 years ago

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Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

3 years ago

Does anyone know the simplest way to find the rate constant in rate law chemistry?

kiruha [24]

Answer:no I don’t sorry

Explanation:

8 0

3 years ago