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2 years ago

What main type of forces must be overcome between br2 when liquid br2 dissolves into ethanol?

Chemistry

1 answer:

Snezhnost [94]2 years ago

8 0

Dipole-induced dipole forces.

In contrast to C2H5OH, which is a very polar molecule, bromine is non-polar and has zero dipole moment. Charges are separated within polar molecules.

When a nonpolar Br2 molecule interacts with a polar C2H5OH molecule, one half of the bromine molecule acquires a charge and the other half acquires an opposing charge. A dipole is created when two adjacent bromine molecules have different charges. Bromine molecules must thus overcome the intermolecular forces—also known as dipole-induced dipole forces—that form in order to dissolve in ethanol.

Find more on bromine at : brainly.com/question/865727

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

Masja [62]

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

Explanation:

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:

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

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

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

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

PLZ HELP I WILL GIVE BRAINLY AND 50 POINTS

Oxana [17]

Answer:

1. State problem

2. Form hypothesis

3. Conduct experiment

4. Interpret data

5. Draw conclusion

Explanation:

4 0

3 years ago

Read 2 more answers

Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueou

Alekssandra [29.7K]

Answer:

[K⁺] = 0.107 M

[OH⁻] = 1.13 × 10⁻⁹ M

Explanation:

600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻

⇒ 0.45 × 600 × 10⁻³

= 0.27 moles of Cu²⁺ and (NO₃)²⁻

450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻

⇒ 0.25 × 450 × 10⁻³

= 0.1125 moles of K⁺ and OH⁻

Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)

Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625

=0.21375 moles which is equal to :

⇒ 0.21375/(( 600+450))× 10⁻³

= 0.21375/1050 × 10⁻³

= 0.20357 M

Given that :

(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)

We know that , Ksp = [Cu²⁺][OH⁻]²

2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²

[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357

[OH⁻] = 1.13 × 10⁻⁹ M

[K⁺] = moles of K⁺ /total volume

[K⁺] = 0.1125 / 1050 × 10⁻³

[K⁺] = 0.107 M

6 0

3 years ago

Consider the model of a sodium atom .Which feature does the model show accurately ?

nignag [31]

It shows that the sodium atom has 11 protons and 12 neutrons, arranged in a ball shape.

8 0

4 years ago

Bubbling ammonia into a solution of bromic acid will produce what salt?

prisoha [69]

<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Ammonium bromate

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

5 0

3 years ago