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Lelu [443]
3 years ago
7

The rate constant for a particular zero-order reaction is 0.075 M s-1. If the initial concentration of reactant is 0.537 M it ta

kes ________ s for the concentration to decrease to 0.100 M.
Chemistry
1 answer:
Anarel [89]3 years ago
8 0

Answer:

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

Explanation:

A zero-order reaction follows the equation:

[A] = [A]₀ - kt

<em>Where [A] is actual reaction of the reactant = 0.100M</em>

<em>[A]₀ the initial concentration = 0.537M</em>

<em>k is rate constant = 0.075Ms⁻¹</em>

<em>And t is time it takes:</em>

<em />

0.100M = 0.537M -0.075Ms⁻¹t

-0.437M = -0.075Ms⁻¹t

5.83s = t

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

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3 years ago
What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide
photoshop1234 [79]

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

3 0
3 years ago
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Explanation:

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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]

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

<u>Explanation:</u>

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:

  • <u>Case 1:</u>

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}

  • <u>Case 2:</u>

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}

  • <u>Case 3:</u>

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^-

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Why does electronegativity decrease from right to left within a period?
Firlakuza [10]

Explanation:

so, as you move down a group on the periodic table, the electronegativity of an element decreases because the increased number of energy levels puts the outer electrons very far away from the pull of the nucleus. electronegativity increases as you move from left to right across a period on the periodic table.

4 0
3 years ago
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