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2 years ago

PLEASE HELP!! What is the concentration [H3O+] of a solution with a pH of 13.2? Is the solution acidic, basic, or neutral?

Chemistry

2 answers:

asambeis [7]2 years ago

8 0

$\\ \tt\leadsto pH=-log[H^+]$

$\\ \tt\leadsto -log[H^+]=13.2$

$\\ \tt\leadsto log[H^+]=-13.2$

$\\ \tt\leadsto [H^+]=antilog(-13.2)=6.3\times 10^{-14}M$

Ilya [14]2 years ago

3 0

The concentration of [H3O+] will be 6.3 x $10^{-14}$ M

Mathematically, pH = -log [H+] or -log [H3O+]

With a pH of 13.2:

-log [H3O+] = 13.2

log [H3O+] = -13.2

[H3O+] = 6.3 x $10^{-14}$ M

More on pH can be found here: brainly.com/question/491373

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3 years ago

Hydrogen reacts with 0.771 g of carbon to form 0.90 g of a compound. What is the mass percent of hydrogen in the compound

andriy [413]

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3 years ago

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =

Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

To calculate the number of moles, we use the equation:

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

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

To calculate the heat of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ

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