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Morgarella [4.7K]
3 years ago
13

Which reaction energy diagram below would be spontaneous? ​

Chemistry
1 answer:
velikii [3]3 years ago
4 0
Answer is D. Both I and III are spontaneous graphs because the change in G < 0.
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Which of the following is true about a spontaneous process? it releases energy, it does not require any external action to begi,
ra1l [238]

Answer: Option (b) is the correct answer.

Explanation:

A spontaneous reaction is defined as the process which tends to occur on its own. It does not require any external factor or force in order to start itself.

For example, when we dissolve KCl in water then potassium chloride being ionic in nature will dissolve on its own. Hence, it will be a spontaneous process.

And, a non-spontaneous reaction is defined as a process for the completion of which we have to provide certain conditions.

Thus, we can conclude that the statement it does not require any external action to begin, is true about a spontaneous process.

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4 years ago
Consider the following information. The lattice energy of CsCl is Δ H lattice = − 657 kJ/mol. The enthalpy of sublimation of Cs
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Answer:

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Explanation:

8 0
4 years ago
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How many moles of ethane (C2H6) would be needed to react with 62.3 grams of oxygen gas?
Black_prince [1.1K]
The combustion of ethane is expressed in the balanced reaction C2H6 + 3.5O2= 2CO2 + 3 H2O. Given the mass of oxygen gas, we get the moles of ethane needed by converting this mass to mole (dividing my 32 g/mol), then multiply by 1/3.5 (stioch ratio) and the molar mass of ethane (30 g/mol). The answer is 16.69 grams ethane. 
6 0
3 years ago
Please answer right this us due today please don't guess
Murljashka [212]
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7 0
3 years ago
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If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em
erica [24]

Answer: Wavelength  associated with the fifth line is 397 nm

Explanation:

E=\frac{hc}{\lambda}

\lambda = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant =10973731.6m^{-1}

n_f = Higher energy level = 7

n_i= Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2

\frac{1}{\lambda}=2.52\times 10^{6}m

\lambda}=3.97\times 10^{-7}m=397 nm      1nm=10^{-9}m

Thus wavelength λ associated with the fifth line is 397 nm

7 0
4 years ago
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