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2 years ago

A water solution is found to have a molar oh- concentration of 3.2 x 10-5. the solution would be classified as :_______.

Chemistry

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

A water solution is found to have a molar oh- concentration of 3.2 x 10-5. the solution would be classified as neutral.

The concentration of hydroxide ions (OH-) is measured by pOH. It is a way of expressing how alkaline a solution is. At 25 degrees Celsius, aqueous solutions with pOH values of 7 or less are neutral, whereas those with pOH values of 7 or more are acidic. The hydrogen ion potential is known as pH. The potential of hydroxide ions is known as pOH. 2. It is a scale used to estimate the hydrogen ion (H+) concentration in the solution. The hydroxide ion (OH-) concentration of the solution is measured using this scale.

pH + pOH = 14

pOH = 3.2x 10-5

[OH-] = 10^(-pOH) =10^(- 3.2x 10-5)

= 0.99

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Answer:

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3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

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3 years ago

A chemist dissolves 484 .mg of pure perchloric acid in enough water to make up 240.mL of solution. Calculate the pH of the solut

sdas [7]

Answer:

1.70

Explanation:

The molar mass of perchloric acid is 100.46 g/mol. The moles corresponding to 484 mg (0.484 g) are:

0.484 g × (1 mol/100.46 g) = 4.82 × 10⁻³ mol

4.82 × 10⁻³ moles are dissolved in 240 mL (0.240 L) of solution. The molar concentration of perchloric acid is:

4.82 × 10⁻³ mol/0.240 L = 0.0201 M

Perchloric acid is a strong monoprotic acid, that is, it dissociates completely, so [H⁺] = 0.0201 M.

The pH is:

pH = -log [H⁺] = -log 0.0201 = 1.70

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3 years ago