All categories

3 years ago

What is the molarity of an H2SO4 solution if 49.0 mL are titrated by 68.4 mL of a .333 M NaOH solution

Chemistry

1 answer:

denis23 [38]3 years ago

8 0

Answer:

0.232M

Explanation:

H2SO4+2NaOH--> 2H2O+Na2SO4

(0.0684L)(.333M NaOH) = 0.0227 moles NaOH (1 mol H2SO4/2 mol NaOH)

= 0.0114 moles H2SO4

M = 0.0114 moles/ 0.0490L solution = 0.232M

* Please text me at 561-400-5105 for tutoring: I can do assignments, labs, exams, etc. :)

You might be interested in

How many protons are found in an atom of platinum (pt)?

Natasha_Volkova [10]

78 protons are found in platinum(pt).

5 0

3 years ago

BRAINLIEST TO FIRST CORRECT ANSWER, THANKS!

Tamiku [17]

Answer:

increased

Explanation:

Consuming a compound increases the concentration. When you increase the concentration, the rate constant for that reaction also increases.

5 0

3 years ago

Which statement is correct? A. pKa is not an indicator of acid strength. B. An acid with a small Ka is stronger than an acid wit

Svetradugi [14.3K]

Answer:

Explanation:

Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

Saying that the smaller the pKa value..the stronger the acid is.

3 0

4 years ago

Read 2 more answers

How many atoms are in 5.00 grams of aluminum foil.

sasho [114]

= 30802.53 im pretty sure if im wrong let me know its the best i can do :/

3 0

4 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 $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago