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Mekhanik [1.2K]
3 years ago
7

Draw the structure for 5-isopropyl-2-methyl-3-nonene

Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Explanation:

In the given molecule of 5-isopropyl-2-methyl-3-nonene their are nine carbon atom attached in straight chain.

  • 3-nonene means that double bond is between the third carbon and forth carbon atom.
  • Isopropyl group (-CH(CH_3)_2) on the fifth carbon atom in the chain
  • Methyl group (-CH_3)on second carbon atom in the chain.

Th structure of the given organic molecule is attached as an image.

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Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+
grin007 [14]

<u>Answer:</u> The standard cell potential of the cell is -0.71 V

<u>Explanation:</u>

The half reactions follows:

<u>Oxidation half reaction:</u>  Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V  ( × 3)

<u>Reduction half reaction:</u>  Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V  ( × 2)

The balanced cell reaction follows:

2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)

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

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

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

E^o_{cell}=-2.37-(-1.66)=-0.71V

Hence, the standard cell potential of the cell is -0.71 V

8 0
3 years ago
A sample of gas occupies a volume of 64.0 mL . As it expands, it does 137.1 J of work on its surroundings at a constant pressure
stiks02 [169]

Answer : The final volume of the gas is, 34.8 L

Explanation :

Formula used :

w=p\times (V_2-V_1)

where,

w = work done = 137.1 J = 35.8 L.atm      (1 L.atm = 101.3 J)

p = external pressure = 783 torr = 1.03 atm    (1 atm = 760 torr)

V_2 = final volume = ?

V_1 = initial volume = 64.0 mL = 0.0640 L      (1 L = 1000 mL)

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

35.8L.atm=1.03atm\times (V_2-0.0640L)

V_2=34.8L

Thus, the final volume of the gas is, 34.8 L

5 0
4 years ago
The above shows a balloon full of gas which has a volume of 120.0 mL
bazaltina [42]

Answer: 128 mL

Explanation: 120mL/300k=v2/320k

7 0
3 years ago
Characteristics of the fungus kingdom (all of them)
Nana76 [90]

- Fungi are "eukaryotic heterotrophs"

- Cell Walls contain a substance called Chitin

- All fungi are multicellular except yeast.

- Fungi are made up of Hyphae

- One hypha is only 1 cell thick

- The most common fungi is the DECOMPOSER

5 0
4 years ago
An animal cell placed in a hypertonic solution will shrink in a process called crenation. An animal cell placed in a hypotonic s
Viefleur [7K]

Answer: A = crenation; B = hemolysis; C = hemolysis, D = crenation; E = neither will occur.

Note: The question is incomplete. The complete question is;

An animal cell placed in a hypertonic solution will shrink in a process called crenation. An animal cell placed in a hypotonic solution will swell and potentially burst in a process called hemolysis. To prevent crenation or hemolysis, an animal cell must be placed in an isotonic solution such as 0.9% (m/v) NaCl or 5.0% (m/v) glucose. This does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes. Hint : A 5.0% (m/v) glucose solution will exert the same osmotic pressure as a red blood cell. Also, a 0.9% (m/v) NaCl solution will exert the same osmotic pressure as a red blood cell. This means that any solution with a higher total solute concentration than either of these isotonic solutions will be hypertonic to the cell. Conversely, any solution with a lower total solute concentration than an isotonic solution will be hypotonic to the cell. A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. Solution A: 3.21% (m/v) NaClSolution B: 1.65% (m/v) glucoseSolution C: distilled H2OSolution D: 6.97% (m/v) glucoseSolution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl

Explanation:

First the concentrations of the NaCl and glucose solutions that are isotonic with the red blood is determined.

Concentration of 0.9% m/v NaCl = 0.9 g/100cm³ = 9 g/dm³

Concentration of 5.0% m/v glucose = 5 g/100cm³ = 50 g/dm³

Concentration of NaCl in Solution A (3.21% (m/v) = 3.21 g/100cm³

= 32.1 g/dm³

Solution A is hypertonic, therefore, crenation will occur.

Concentration of 1.65% (m/v) glucose Solution = 1.65 g/100cm³ = 16.5 g/dm³

Solution B is hypotonic, therefore, hemolysis will occur.

Solution C,  Distilled water contains no dissolved solutes. The solution is hypotonic, therefore, hemolysis will occur.

Concentration of 6.97% (m/v) glucose solution = 6.97 g /100cm³

= 69.7 g/dm³

The solution is hypertonic. Therefore, crenation will occur.

Concentrations of 5.0% (m/v) glucose and 0.9%(m/v) NaCl are 50 g/dm³ and 9 g/dm³ respectively. Both solutions are isotonic to the red blood cell. Therefore, neither crenation nor hemolysis will occur.

5 0
3 years ago
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