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mel-nik [20]
3 years ago
7

What is the [h3o+] in a solution with a poh = 5.20?

Chemistry
1 answer:
Pepsi [2]3 years ago
7 0
POH scale is used to measure how alkaline a solution is
pOH can be calculated as follows
pOH = - log [OH⁻]
pOH is 5.20 and using this the pH can be calculated
pOH + pH = 14
pH = 14 - pOH 
pH = 14 - 5.20 = 8.80
pH scale is used to measure how acidic a solution is
pH = -log [H₃O⁺]
[H₃O⁺] = antilog (-8.80)
[H₃O⁺] = 1.6 x 10⁻⁹ M
answer is [H₃O⁺] = 1.6 x 10⁻⁹ M 
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Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
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<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

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

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

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

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

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