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laila [671]
3 years ago
13

Under what conditions can a non-spontaneous reaction be made spontaneous?

Chemistry
2 answers:
Kitty [74]3 years ago
7 0
It can be c I think tho
Gre4nikov [31]3 years ago
6 0

Answer:

A. When an endothermic reaction has a positive change in entropy and the temperature is greatly increased

Explanation:

An spontaneous reaction is one that occurs without having been provoked by an external influence. The non-epontaneous reaction, on the other hand, is one that occurs only when they are induced in the opposite direction.

Most of the non-spontaneous reactions are endothermic, that is, they occur with the absorption of energy. An example of a non-spontaneous reaction occurs when a metal needs to be heated to increase its temperature. The heating of a metal is very difficult to occur naturally, without an external factor that provokes the reaction.

However, a non-spontaneous reaction can become spontaneous depending on the entropy of the system. Entropy in an isolated system tends to increase when an spontaneous reaction occurs. Therefore, in order for a non-spontaneous reaction to become spontaneous it is necessary that an endothermic reaction has a positive change in entropy and the temperature increases a lot.

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Why is it better to conduct an experiment more than once?
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The first reason to repeat experiments is simply to verify results. Different science disciplines have different criteria for determining what good results are. Biological assays, for example must be done in at least triplicate to generate acceptable data. Science is built on the assumption that published experimental protocols are repeatable.


2)      The next reason to repeat experiments is to develop skills necessary to extend established methods and develop new experiments. “Practice make perfect” is true for the concert hall and the chemical laboratory.


3)      Refining experimental observations is another reason to repeat. Maybe you did not follow the progress of the reaction like you should have.


4)      Another reason to repeat experiments is to study and/or improve them in way. In the synthetic chemistry laboratory, for example, there is always a desire to improve the yield of a synthetic step. Will certain changes in the experimental conditions lead to a better yield? The only way to find out is to try it! The scientific method informs us that it is best to only make one change at a time.


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8 0
3 years ago
What do you call a bond that forms when electrons are transferred from one atom to another?
AlekseyPX
It is called an ionic bond
4 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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The value of the rate constant for a gas phase reaction can be changed by increasing the A. temperature of the reaction vessel.
prohojiy [21]

Answer:

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Explanation:

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Convert 4.0 cm to mm.
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Answer: 40mm

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