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katrin [286]
3 years ago
9

If the hydroxide ion concentration of an aqueous solution is 3.0 x 10-4, what is the [H+] in the solution? Is the solution acidi

c, basic, or neutral?
Chemistry
1 answer:
Volgvan3 years ago
6 0

Answer:

[H+] is  3.31 × 10^-11 M

Solution: Basic

Explanation:

In the question we are given;

Concentration of Hydroxide ion [OH-] as 3.0 × 10^-4 M

We are required to determine the [H+] and whether the solution is acidic, basic or neutral.

We know that;

pOH + pH = 14

and that;

pOH = - log [OH-]

Therefore;

pOH = - log ( 3.0 × 10^-4)

       = 3.52

But;

pH = 14 - pOH

Thus;

pH = 14 - 3.52

     = 10.48

But, we know that;

pH = - log[H+]

Therefore;

H+ = Antilog -pH

    = Antilog -10.48

    = 3.31 × 10^-11

Thus, the concentration of H+ ions [H+] is  3.31 × 10^-11 M

Also since the pH is 10.48, then the solution is basic

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[2.8.8]^-  (answer  B)

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3 years ago
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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

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

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

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

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

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

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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Answer:

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Answer:

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Explanation:

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dangina [55]

Answer:

Explanation:

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