Which of the following could be a long-term health effect of chemical pesticide use?

How many moles of Cu2S can be formed from 11 moles of HCl?

natima [27]

Answer:

hi

Explanation:

hi

4 0

2 years ago

Which of the following hypotheses cannot be tested using the scientific method

Arturiano [62]

I believe it's answer #3. Logically, at least.

You can test #1 through trial and error.
You can experiment #2 also through trial and error.
You cannot test #3 through trial and error, because that would be catastrophic.
You can test #4 through a survey and individual study and data collection.

7 0

3 years ago

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A 100g sample of a metal was heated to 100oC and then quickly transferred to an insulated container holding 100g of water at 22o

Lady_Fox [76]

Answer:

B) The metal temperature changed more than the water temperature did, but the metal lost

the same amount of thermal energy as the water gained.

Explanation:

Heat capacity or thermal capacity is defined as the amount of heat required by a given mass of a material to raise its temperature by one unit which means that the heat capacity of the water, that is the quantity of heat required to cause a rise from 22°C to 35°C that is a rise of 13°C is the quantity of heat that caused the drop in temperature of the metal from 100°C to 35°C a change of 65°C

The water has more capacity to absorb heat or a higher heat capacity than the metal

However, the first law of thermodynamics states that energy is neither created nor destroyed, but it changes from one form to another. In this case, the thermal energy lost by the metal is the same as the thermal or heat energy gained by the water

4 0

2 years ago

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

Terrrrrrrrrrrrrrrrrrrrrrrr?

FrozenT [24]

what is your question?

6 0

2 years ago

Read 2 more answers