The diagram shows an early model of the atom. It was suggested by JJ Thompon. The atom was thought to be positively charged

8. What group numbers are the transition elements in?

Anna007 [38]

The 38 elements in groups 3 through 12 of the periodic table are called "transition metals". As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat. The interesting thing about transition metals is that their valence electrons, or the electrons they use to combine with other elements, are present in more than one shell. This is the reason why they often exhibit several common oxidation states. There are three noteworthy elements in the transition metals family. These elements are iron, cobalt, and nickel, and they are the only elements known to produce a magnetic field I HOPE THIS HELP:)

3 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

What are land plants used for?

kotegsom [21]

Answer:

Land plants are also known as Embryophytes. Embryophytes are complex multicellular eukaryotes with specialized reproductive organs. The name derives from their innovative characteristic of nurturing the young embryo sporophyte during the early stages of its multicellular development within the tissues of the parent gametophyte.

Explanation:

4 0

3 years ago

Read 2 more answers

Which statement about atoms and molecules is

Pavel [41]

Elements always exist as pair of atoms called molecules .

Explanation:-

The material which has only one types of similar atoms called element .
Ex:-Sodium,Carbon etc

8 0

3 years ago

Read 2 more answers

The kb of hypochlorous acid is 3. 0×10^–8 at 26. 0 C. What is the percent of ionization of hypochlorous?

astraxan [27]

The acid dissociation constant (Ka) defines the difference between a weak and a strong acid. The % ionization of hypochlorous acid is 0.14%.

<h3>What is the acid dissociation constant?</h3>

The acid dissociation constant is used to define the ionization constant of an acidic substance. It gives the quantitative measurement of the strength.

The ICE table is attached to the image below.

The acid dissociation constant (Ka) for the reaction is,

Ka = [H⁺][ClO⁻] ÷ [HClO]

= a² ÷ (0.015 - a)

= 3.0 x 10⁻⁸

Now, a² + 3.0 x 10⁻⁸ a - 4.5 × 10⁻¹⁰ = 0

So, a = 2.210 × 10⁻⁵

Solving further,

[H+] = a = 2.210 × 10⁻⁵ M

The percent ionization is calculated as,

[H+] ÷ [HClO] × 100

= 2.210 × 10⁻⁵ M ÷ 0.015 × 100

= 0.14 %

Therefore, 0.14 % is the percentage of hypochlorous ionization.

Learn more about acid dissociation constant here:

brainly.com/question/22668939

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Your question is incomplete, but most probably your full question was, The ka of hypochlorous acid (HClO) is 3.0 x 10⁻⁸ at 25.0°C. What is the % of ionization of hypochlorous acid in a 0.015 aqueous solution of HClO at 25.0C?

6 0

1 year ago