answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
julsineya [31]
2 years ago
5

Draw the structural formula of 4,5-diisopropylnonane.

Chemistry
1 answer:
mamaluj [8]2 years ago
3 0
Following the IUPAC rules, the longest carbon chain should consist of 9 carbons because of the word 'nonane'. From there, attach two isopropyl (3-carbon chains) to the 4th and 5th carbon starting either from the left or right of the parent carbon chain. The structural formula is shown in the picture attached.

You might be interested in
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

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}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
Water treatment plants commonly use chlorination to destroy bacteria. a byproduct is chloroform (chcl3), a suspected carcinogen
antiseptic1488 [7]
<span>100. ppb of chcl3 in drinking water means  100 g of CHCl3 in 1,000,0000,000 g of water

Molarity, M

M = number of moles of solute / volume of solution in liters

number of moles of solute = mass of CHCl3 / molar mass of CHCl3

molar mass of CHCl3 = 119.37 g/mol

number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol

using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters

M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer

Molality, m

m = number of moles of solute / kg of solvent

number of moles of solute = 0.838

kg of solvent = kg of water = 1,000,000 kg

m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer

mole fraction of solute, X solute

X solute = number of moles of solute / number of moles of solution

number of moles of solute = 0.838

number of moles of solution = number of moles of solute + number of moles of solvent

number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles

number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles

X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer

mass percent, %

% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =

% = 10 ^ - 6 % <------- answer
</span>
7 0
2 years ago
Read 2 more answers
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2
LekaFEV [45]
Molar mass CaCl₂ = 111.0 g/mol

number of moles:

n = mass of solute / molar mass

n = 85.3 / 111.0

n = 0.7684 moles of CaCl₂

M = n / V

0.788 M = <span>0.7684 / V
</span>
V = 0.7684 / 0.788

V = 0.97512 L

hope this helps!


3 0
2 years ago
Trichloromethane consists of 10.05% carbon. 0.83% hydrogen and 89.12% chlorine. if uts relative molecular mass is 119.5g . find
Delicious77 [7]

Answer:

C24H2CL213

Explanation:

C

(10.05/100)x119.5

=12x2=24

H

(0.083/100)x119.5

=1x2=2

Cl

(89.12/100)x119.5

=106.5x2=213

4 0
2 years ago
A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant
VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is,             Volume \propto Temperature

Hence, it is given that V_{1} is 3.50 liters, T_{1} is 20 degree celsius, and T_{2} is 100 degree celsius.

Therefore, calculate V_{2} as follows.

                           \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

                           \frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}

                                V_{2} = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0
2 years ago
Read 2 more answers
Other questions:
  • What is the percentage concentration by mass of a solution that contains 0.098 kg of H2SO4 in 500.0 g of H2O?
    14·2 answers
  • A sample of hydrogen gas was collected over water at 21ºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m
    10·1 answer
  • given that the molar mass of Na2SO4 × nH2O is 322.1 g/mol calculate the number of moles of water in this hydrate
    11·1 answer
  • Select the most acidic solution from these options. select the most basic (alkaline) solution from these options. select the lea
    9·2 answers
  • Match each decimal number to the correct scientific notation. 30,700 0.000307 0.00000307 3,070,000 3.07 × 10-6 arrowRight 3.07 ×
    12·2 answers
  • Compound A, C6H12 reacts with HBr/ROOR to give compound B, C6H13Br. Compound C, C6H14, reacts with bromine and light to produce
    5·1 answer
  • 3. For the reaction: 2X + 3Y 3Z, the combination of 2.00 moles of X with 2.00
    6·1 answer
  • A radioisotope is placed near a radiation detector, which registers 80 counts per second. Eight hours later, the detector regist
    11·1 answer
  • A mercury mirror forms inside a test tube as a result of the thermal decomposition of mercury(ii) oxide: 2 hgo(s) → 2 hg(l) + o2
    8·1 answer
  • onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!