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3 years ago

Energy conversion means reducing energy use True or false

Chemistry

1 answer:

I am Lyosha [343]3 years ago

3 0

False, energy conversion just means the energy is going to be used by another force

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zzz [600]

Answer:

soorry i ddsmfcj know

Explanation:

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3 years ago

I NEED HELP ASAP WILL MARK BRAINLIEST!!!

andreev551 [17]

The answer is b.) Levi crosses a horse with a donkey to a create a mule. The mule is stronger than both of its parents.

7 0

3 years ago

What is the empirical formula of a compound containing 40.0% sulfur and 60.0% oxygen by mass?

gizmo_the_mogwai [7]

in order to determine empirical formula we have to determine the mole ratio of the given elements

Let the total mass of the compound is 100g

as given that the compound has 40% sulfur , so mass of sulfur = 40g

as given that the compound has 60% oxygen, so mass of oxygen = 60g

let us calculate the moles of each element

Moles of sulfur = mass / atomic mass = 40 / 32 = 1.25

moles of oxygen = mass / atomic mass = 60/ 16 = 3.75

In order to get simple ratio of moles we will divide both the moles with least number of moles which is 1.25

moles of sulfur = 1.25 / 1.25 = 1

moles of oxygen = 3.75 /1.25 = 3

So empirical formula will be SO₃

3 0

3 years ago

HELP! Give the molar mass of Copper (II) Hydroxide.

yarga [219]

Answer:

97.561 g/mol

Explanation:

8 0

3 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