4 On a bright, sunny day, which time of day can

Mengapa garam NaNO2 dapat membirukan kertas lakmus merah? Jelaskan dengan reaksi kimianya!

Reika [66]

Can't understand this. English Please! :) :)

8 0

3 years ago

You have two solid substances that look the same. What measurements would you take and which tests would you preform to determin

ASHA 777 [7]

Answer:

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5 0

2 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

Question 10

bixtya [17]

The reaction that should be followed is
Na2SO4 + Ca(NO3)2 --> CaSO4 + 2NaNO3
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (1.54×10−2 mol / L) = 1.155x10-3 mol

mol Ca(NO3)2 = 0.075 L (1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2

so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 %

5 0

3 years ago

#1: The following reaction occurs in a voltaic cell: Cu2+ + Zn --> Zn2+ + Cu. Which of the following reactions occurs at the

Rudiy27

In a voltaic cell there are two electrodes: anode and cathode.

On the cathode, the reaction that occurs is reduction, which means that the metal gains electrons.

On the anode the reaction is oxidation, which means that the metal loses electrons.

In this voltaic cell the reaction on the anode is:

Zn --> Zn 2+ + 2e-, i.e. option C.

Answer: option C.

5 0

3 years ago