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3 years ago

What is the approximate value of the ccc bond angle in ch3ch2ch2oh

Chemistry

1 answer:

Fynjy0 [20]3 years ago

6 0

Answer:

The value of the carbon bond angles are 109.5 °

Explanation:

CH3CH2CH2OH = propanol . This is an alcohol.

All bonds here are single bonds.

Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will create four hybrid orbitals with similar characteristics.

Sp3- types have angles of 109.5 ° between the carbon - atoms.

This means that the value of the carbon bond angles are 109.5 °

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Kobotan [32]

Answer:

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Explanation:

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3 years ago

I need help doing this please. It’s an image

Ne4ueva [31]

Answer:

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3 years ago

A sample of nitrogen gas is produced in a reaction and collected under water in a graduated cyliner. The temperature is 26.3 oC

Rudiy27

The mass of nitrogen collected is mathematically given as

M-N2=0.025gram

<h3>What is the mass of nitrogen collected?</h3>

Question Parameters:

A sample weighing 2.000g

the liberated NH3 is caught in 50ml pipeful of H2SO4 (1.000ml = 0.01860g Na2O).

T=26.3c=299.3K

Pressure=745mmHg=745torr

Pressure of N2=745-25.2=719.8torr

Generally, the equation for the ideal gas is mathematically given as

PV=nRT

Therefore

719.8/760=45.6/1000=n*0.0821*299.3

n=0.00176*14

In conclusion, the Mass of N2

M-N2=0.00176*14

M-N2=0.025gram

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8 0

2 years ago

g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular

Whitepunk [10]

Answer:

Where the products are H2O and Ba(NO3)2

Explanation:

A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.

The balanced reaction is:

Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2

Where the products are H2O and Ba(NO3)2

4 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