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3 years ago

Why carbon dioxide is much more soluble in alkaline solutions

Chemistry

2 answers:

disa [49]3 years ago

8 0

Answer:

because co2 dissolve by water

rewona [7]3 years ago

3 0

Answer:

Because CO2 dissolved in water is acidic: it forms carbonic acid. The second equilibrium below strongly favors dissolved CO2

Hope it will help you

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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

Mercury can be obtained by reacting mercury(I) sulfide with calcium oxide. How many grams of calcium oxide are needed to produce

OLga [1]

First, 55 g of Hg is 3.65 moles because one mole of Hg has a molar mass of 200.59

Then, the mole ratio of Hg to CaO is 8:4 or 2:1. SO we divide 3.65 by 2 to get 1.82 moles of CaO

This is the same as 102.06 grams because one mole of CaO has a molar mass of 56.0774

Hope this helps!

4 0

3 years ago

Describe how a positive result for the Baeyer Test will look. What is being oxidized? What is being reduced?

Ksivusya [100]

Answer:

This reaction is sometimes referred to as the Baeyer test. Because potassium permanganate, which is purple, is reduced to manganese dioxide, which is a brown precipitate, any water‐soluble compound that produces this color change when added to cold potassium permanganate must possess double or triple bonds.

Explanation:

5 0

3 years ago

Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm

ivolga24 [154]

Answer:

$O_{2}$ and $NO_{2}$

Explanation:

For a given system at constant temperature, the number of moles of gas present in the system is proportional to the product of the system pressure and volume. Therefore, we have:

NO: 6 L * 0.7 atm = 4.2 L*atm

O: 1.5 L* 2.5 atm = 3.75 L*atm

For the given system based on a balanced chemical equation:

2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e. $P_{NO}$ = 0

For the product: 2.70 L*atm NO produced 2.70 L*atm $NO_{2}$

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of $NO_{2}$ = (2.70 L*atm $NO_{2}$ ) / (7.5 L) = 0.36 atm $NO_{2}$

Similarly for oxygen gas:

3.75 L*atm - 1.35 L*atm = 2.40 L*atm oxygen gas remaining

Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm

Thus, the gases present at the end of the experiment are $O_{2}$ and $NO_{2}$

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What is a part of every cell

Dafna11 [192]

Most accurate cell membrane

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Why carbon dioxide is much more soluble in alkaline solutions​

Why carbon dioxide is much more soluble in alkaline solutions