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3 years ago

Please help with this , thanks you guys very much

Chemistry

1 answer:

ahrayia [7]3 years ago

4 0

Answer:

I am giving u some explanation related this question pls see that 

it may help u

Explanation:

The pH scale measures how acidic an object is. Objects that are not very acidic are called basic. The scale has values ranging from zero (the most acidic) to 14 (the most basic). As you can see from the pH scale above, pure water has a pH value of 7. This value is considered neutral—neither acidic or basic. Normal, clean rain has a pH value of between 5.0 and 5.5, which is slightly acidic. However, when rain combines with sulfur dioxide or nitrogen oxides—produced from power plants and automobiles—the rain becomes much more acidic. Typical acid rain has a pH value of 4.0. A decrease in pH values from 5.0 to 4.0 means that the acidity is 10 times greater.

How pH is Measured

There are many high-tech devices that are used to measure pH in laboratories. One easy way that you can measure pH is with a strip of litmus paper. When you touch a strip of litmus paper to something, the paper changes color depending on whether the substance is acidic or basic. If the paper turns red, the substance is acidic, and if it turns blue, the substance is basic

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Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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How do the fusion reaction in the sun compare to the fusion occurring in large atars and supernovas

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The correct answer is D).

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PIT_PIT [208]

All of them are agree :)

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