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3 years ago

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6. Menthol is a member of the terpene family of natural products. It exists in a (1R, 2S, 5R) form and a (IS, 2R, 5S) form. Are

these two compounds enantiomers or diastereomers?

1 answer:

Usimov [2.4K]3 years ago

5 0

Here ya go if you need the link we’re I found this answer key https://studylib.net/doc/8708211/exam-2---chemistry

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The concentration of a biomolecule inside a rod-shaped prokaryotic cell is 0.0027 m. calculate the number of molecules inside th

mash [69]

Answer : The number of molecules inside the cell is, $8.08\times 10^6$

Solution :

First we have to calculate the volume of biomolecule cell.

Volume = Area of the base of the cell × length of the cell

$V=\pi r^2\times h$

where,

V = volume of the biomolecule cell

r = radius of the cell = $\frac{Diameter}{2}=\frac{1.2}{2}=0.6\mu m$

h = length of the cell = $4.4\mu m$

Now put all the given values in the above volume formula, we get

$V=\frac{22}{7}\times (0.6\mu m)^2\times (4.4\mu m)=4.978\mu m^3=4.978\times 10^{-15}L$

conversion : $(1\mu m^3=10^{-15}L)$

Now we have to calculate the moles of a biomolecule of the cell.

$Molarity=\frac{Moles}{Volume}\\\\Moles=Molarity\times Volume=(0.0027mole/L)\times (4.976\times 10^{-15}L)=0.01343\times 10^{-15}moles$

Now we have to calculate the number of molecules inside the cell.

$\text{Number of molecules}=Moles\times (6.022\times 10^{23})\\\\\text{Number of molecules}=(0.01343\times 10^{-15})\times (6.022\times 10^{23})=8.08\times 10^6$

Therefore, the number of molecules inside the cell is, $8.08\times 10^6$

4 0

3 years ago

A principal quantum number refers to:

lys-0071 [83]

The size of the orbital and the energy level an electron is placed in

5 0

3 years ago

For fun<br><br> need full details to win<br> would you rather be a man or a women and why

ollegr [7]

Answer:

i rather be a man so i have to go through the same pain every month

Explanation:

3 0

3 years ago

Read 2 more answers

Multiple choice-- please help!

kvasek [131]

The rate law for the reaction : r=k.[A]²

<h3>Further explanation</h3>

Given

Reaction

A ⟶ B + C

Required

The rate law

Solution

The rate law is a chemical equation that shows the relationship between reaction rate and the concentration / pressure of the reactants

For the second-order reaction it can be:

1. the square of the concentration of one reactant.

$\tt r=k[A]^2$

2. the product of the concentrations of two reactants.

$\tt r=k[A][B]$

And the reaction should be(for second order) :

2A ⟶ B + C

Thus, for reaction above (reactant consumption rate) :

$\tt r=-\dfrac{\Delta A}{2\Delta t}=k[A]^2$

6 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as 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:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago