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DanielleElmas [232]
3 years ago
10

The volume of 160. g of CO initially at 273 K and 1.00 bar increases by a factor of two in different processes. Take CP,m to be

constant at the value 29.14 Jmol−1K−1 and assume ideal gas behavior. The temperature of the surroundings is 273 K.
A) Calculate ΔSsurroundings in an adiabatic reversible expansion.

B) Calculate ΔS in an adiabatic reversible expansion.

C) Calculate ΔStotal in an adiabatic reversible expansion.

D) Calculate ΔSsurroundings in an expansion against Pexternal = 0.

E) Calculate ΔS in an expansion against Pexternal = 0.

F) Calculate ΔStotal in an expansion against Pexternal = 0.

G) Calculate ΔSsurroundings in an isothermal reversible expansion.

H) Calculate ΔS in an isothermal reversible expansion.

I) Calculate ΔStotal in an isothermal reversible expansion.

Determine what processes are spontaneous.

1) adiabatic reversible expansion 2) expansion against Pexternal =0 3)isothermal reversible expansion
Chemistry
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

Explanation:

w eFedfweF edf SEDFAsFGFSDSFG BAEWRDA G

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Name at least four other gases in the atmosphere besides oxygen and nitrogen
nikitadnepr [17]
Carbon dioxide (CO2)
Argon (Ar)
Hydrogen (H)
Helium (He)
8 0
3 years ago
Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+
eimsori [14]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

X\rightarrow X^{n+}+ne^-

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

  • <u>Option a:</u>  Ag^+

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

  • <u>Option b:</u>  Pb

This metal can easily get oxidized to Pb^{2+} ion and the standard oxidation potential for this is 0.13 V

Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V

  • <u>Option c:</u>  H_2

This metal can easily get oxidized to H^{+} ion and the standard oxidation potential for this is 0.0 V

H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V

  • <u>Option d:</u>  Ag

This metal can easily get oxidized to Ag^{+} ion and the standard oxidation potential for this is -0.80 V

Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V

  • <u>Option e:</u>  Mg^{2+}

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive E^o potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0
3 years ago
In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 &gt; 4NO + 6H2O a. Which reactant is the limi
cluponka [151]

Answer:

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

7 0
3 years ago
Help?
Ahat [919]

Answer:

KClO_3

Explanation:

Hello!

In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:

m_O=22.9g-7.33g-6.65g=8.92g

Next, we compute the moles of each element:

n_K=\frac{7.33g}{39.9g/mol}= 0.184mol\\\\n_{Cl}=\frac{6.65g}{35.45g/mol}=0.188mol \\\\n_O=\frac{8.92g}{16.00g/mol} =0.5575mol

Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:

K=\frac{0.184}{0.184}=1.0 \\\\Cl=\frac{0.187}{0.184}=1.0\\\\O=\frac{0.5575}{0.184}  =3.0

Therefore, the empirical formula is:

KClO_3

Regards!

3 0
2 years ago
Help!!will give brainliest
Aleksandr-060686 [28]

Answer:

As a pyramid gets taller, it gets smaller. In the Linnaean classification system, this happens as well, with each category growing smaller.

Explanation:

7 0
3 years ago
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