All categories

2 years ago

3,7-dibromo-4,6-dichloro oct-5-ene-1,2-diol. --- show its structure by photo.

Chemistry

2 answers:

valkas [14]2 years ago

6 0

Explanations:- Looking at the given name, the parent chain name is Oct and Oct stands for eight carbons.

The suffix -ene at number 5 stands for a double bond between carbon number 5 and 6.

Bro stands for Br. Since the name has dibromo means there are two Br at number 3 and 7 carbons.

Similarly, there are two Cl at number 4 and 6 carbons.

Suffix -ol is used for OH group. The given name has 1,2-diol means there are two OH groups at number 1 and 2 carbons.

So, to draw the structure of the compound we put 8 carbons in a chain and then but a double bond between number 5 and 6 carbons. Put OH at number 1 and 2 carbons, put Br at number 3 and 7 carbons and Cl at number 4 and 6 carbons. If any of the carbons octet (four bonds) is still incomplete then we attach H atoms to it.

So, the structure of the compound is as given below:

VikaD [51]2 years ago

4 0

Refer to the attachment for answer.

SOME EXTRA INFORMATION:

Bromo is used for Bromine (Br)

Chloro is used for Chlorine(Cl)

'ene' represents double bond between carbon atoms.

'ol' is used for alcohol

HOPE IT IS USEFUL

You might be interested in

Please help, I'm struggling with stoichiometry

rusak2 [61]

Answer:

a) HNO3

b) 26.8g (3 s.f.)

c) 1.29g (3 s.f.)

Please see the attached pictures for full solution.

To balance an equation, ensure that the number of atoms for each element is the same on both sides.

3 0

3 years ago

How does a rock<br>Wither? Explain.

lesantik [10]

First, it combines with carbon dioxide in the soil to form a weak acid called carbonic acid. ... Carbonic acid slowly dissolves away minerals in rock, especially the carbonate minerals that make up limestone and marble. The weak acid decomposes the insoluble rock into watersoluble products that move into the groundwater.

3 0

2 years ago

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

A worker at a nuclear power plant has noticed that his dosimeter alarm has

Sphinxa [80]

Answer:

Radiation is being released from the reactor.

Explanation:

( A P E X )

5 0

2 years ago