All categories

4 years ago

The compounds 2-butanol and 2-butene both contain

2 answers:

4 0

The answer is (3) carbon atoms. For 2-butanol, it contains single bonds only and with C, O, H atoms. For 2-butene, it contains single bonds and double bonds both. And contains C, H atoms.

RoseWind [281]4 years ago

3 0

Answer:

(3) carbon atoms

Explanation:

2- butanol is a secondary alcohol containing C,H and O with a molecular formula: C4H10O and a structural formula: CH3CH2CH(OH)CH3

2- butene is an unsaturated hydrocarbon i.e. an alkene containing C and H only with a molecular formula: C4H8 and a structural formula:

CH3CH2CH=CH2

Based on the given options, the only common ground between 2-butanol and 2-butene is that both contain carbon atoms.

You might be interested in

Jeffry was wondering why when he made breakfast, that after he was done cooking, the mass of everything he used was not the same

12345 [234]

Answer:

equivalent exchange forces cancel out but the substances are affected around the area

Explanation:

4 0

3 years ago

HELP WITH A THESIS STATEMENT

lina2011 [118]

The second thesis statement is perfect. It supports the claim and presents main idea.

4 0

3 years ago

Elements that belong to the same period/row!!

-BARSIC- [3]

Answer:

Each row is called a period. Each column is called a group or family.

Explanation:

7 0

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

8 0

3 years ago

Can any one please help me to answers this question i don't get it. thank in advance.

alex41 [277]

First, lets balance the reaction equation:
4Fe + 3O₂ → 2Fe₂O₃
It is visible form the equation that 4 moles of Fe require 3 moles of O₂
Molar ratio Fe/O₂ = 4/3 = 1.33
Molar ratio O₂/Fe = 3/4 = 0.75
Now, we check the molar ratios present:
Fe/O₂ = 6.8/8.9 = 0.76
O₂/Fe = 1.31

Thus, Iron is the limiting reactant because its ratio is not being fulfilled while the ratio of O₂ is surpassed.

8 0

3 years ago