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3 years ago

Indicate whether the following are acidic, basic, or neutral solution

1 answer:

5 0

You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:

A. [H2O+]=6.0x10^-12
basic

B. [H3O+]=1.4x10^-9
basic

C. [OH-]=5.0x10^-12
acidic

D. {OH-]=3.5x10^-10
acidic

Hope this answers the question.

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A bowling ball and basketball collide, what direction do they go and how is energy transferred?

olga nikolaevna [1]

Answer:

When the batter hits the ball, there is a force applied, and energy is transferred. The ball will move in the direction the force is pushing it. If two objects collide, energy will be transferred between both, and there will be a change in motion.

Explanation:

8 0

2 years ago

Metals can be highly purified using electrolysis. Which equation best reflects the overall process of purifying copper?

makvit [3.9K]

Answer:

A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode

Explanation:

Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it. The resulting changes at the anode and cathode are given by the equation:

cathode: Cu²⁺ + 2e⁻ ⇒ Cu

anode: Cu ⇒ Cu²⁺ + 2e⁻

3 0

3 years ago

Read 2 more answers

2a. Write down the names of the new substances formed when each of the following substance reacts with dilute hydrochloric acid.

Naddika [18.5K]

Answer:

i) Dilute hydrochloric acid will react with Ammonia to form ammonia salt.

ii) dilute hydrochloric acid will react with soduim hydroxide to form sodium chloride and water

iii) dilute hydrochloric acid will react with calcuim carbonate to form Calcium chloride, Carbon dioxide and water.

Explanation:

CHEMICAL EQUATIONS :

I) HCL + NH3 = NH4Cl

ii) HCL+ NaOH = NaCl + H2O

iii) HCL + CaCo = CaCl2 + CO2 + H2O

7 0

2 years ago

Read 2 more answers

Please help ————//-//

grandymaker [24]

Probably b because soil can’t be the exact same and so that cancels out a and d

6 0

3 years ago

Read 2 more answers

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago