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agasfer [191]
2 years ago
10

Calculate the pH for the following solutions and indicate whether each solution is acidic or basic.

Chemistry
2 answers:
Nadya [2.5K]2 years ago
6 0
The answer would for sure be 178 because acid & base make that amount so yea!! I’m talllyyy right
garik1379 [7]2 years ago
5 0

\\ \rm\Rrightarrow pH=-log[H^+]

\\ \rm\Rrightarrow pH=-log10^{-2}

\\ \rm\Rrightarrow pH=2

Acidic

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

<u>Explanation:</u>

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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