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jeka57 [31]
3 years ago
12

For resonance forms of a molecule or ion, ________.

Chemistry
1 answer:
Sergio [31]3 years ago
6 0
<span>For resonance forms of a molecule or ion, the observed structure is an average of the resonance forms. Sometimes it also depends on how the ion or molecule interact with neighboring molecule or ion. In physics, resonance</span><span> is described as when a vibrating system or external force drives another system to oscillate with greater amplitude at a specific preferential frequency. </span><span>It is not that one structure always corresponds to the observed structure because they will not exist if it is drawn. It is not the same atoms that do not need to be bonded to each other of the resonance structure because it is describing the migration of the atoms not the species itself. </span>
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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
Galina-37 [17]

Answer:

Kp = 41.53

Kc = 1.01

Explanation:

To calculate the equilibrium constant in terms of pressure, what we simply do is to use the equilibrium pressure raised to the power of the number of moles. What we are saying in essence is this:

Kp = [NOCl]^2/[NO]^2[Cl]

Kp= [0.25]^2/[0.174][0.093]^2 = 41.53

Kp = Kc (RT)^Dn

Hence, Kc = Kp/[RT]^(delta n )^-1

n = sum of the number of moles of products minus the sum of the number of moles of reactants= 2-3 = -1 in this case

Kc = 41.53/(0.0821 * 500)^1

Kc = 1.01

7 0
3 years ago
Read 2 more answers
Don’t waste your time answering with links because I won’t even open it. Plz just give a normal answer. Thanks.
Norma-Jean [14]

im really sorry I need points so I’m answering this again really sorry!

3 0
3 years ago
Read 2 more answers
A) Calculate the standard free-energy change at 25 ∘C for the following reaction:
Genrish500 [490]

Answer:

A) ΔG° = -3,80x10⁵ kJ

B) E° = 2,85V

Explanation:

A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:

Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)

The ΔG° of reaction is:

ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>

Where:

ΔGFe(s): 0kJ

ΔGMg²⁺(aq): -458,8 kJ

ΔGFe²⁺(aq): -78,9 kJ

ΔGMg(s): 0kJ

Replacing in (1):

ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)

<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>

B) For the reaction:

X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)

ΔG° = ΔH° - (T×ΔS°)

ΔG° = -629000J  - (298,15K×-263J/K)

ΔG° = -550587J

As ΔG° = - n×F×E⁰

Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential

Replacing:

-550587J = - 2mol×96485J/Vmol×E⁰

<em>E° = 2,85V</em>

I hope it helps!

3 0
3 years ago
How do we use biomass as an energy source?
kherson [118]
Biomass<span> contains stored </span>energy<span>. That's because plants absorb </span>energy<span> from the sun through the process of photosynthesis. When </span>biomass<span> is burned, this stored</span>energy<span> is released as heat. Burning </span>biomass<span> releases carbon dioxide

</span>
6 0
4 years ago
The electrolysis of water forms H2 and O2. 2H2O 2H2 + O2 What is the percent yield of O2 if 10.2 g of O2 is produced from the de
BigorU [14]

<u>Answer:</u> The percent yield of oxygen gas is 67.53 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of water = 17.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol

For the given chemical equation:

2H_2O\rightarrow 2H_2+O_2

By Stoichiometry of the reaction:

2 moles of water produces 1 mole of oxygen gas

So, 0.944 moles of water will produce = \frac{1}{2}\times 0.944=0.472moles of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.472 moles

Putting values in equation 1, we get:

0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g

To calculate the percentage yield of oxygen gas, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of oxygen gas = 10.2 g

Theoretical yield of oxygen gas = 15.104 g

Putting values in above equation, we get:

\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%

Hence, the percent yield of oxygen gas is 67.53 %.

7 0
4 years ago
Read 2 more answers
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