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2 years ago

Which metal in Period 5 is very reactive and has two valence electrons in each atom?

2 answers:

8 0

The answer is Strontium(Sr). The reactive increase from right to left. And this element has two valence electrons. So Rb is not correct. Then the very reactive metal is Sr.

irakobra [83]2 years ago

8 0

Answer: The correct answer is Strontium.

Explanation:

Reactivity of an element is defined as the tendency to loose or gain electrons. Reactivity decreases as we move from left to right in a period and it increases as we move from up to down in a group.

A metal which belongs to Period 5 and has 2 valance electrons is Strontium. This element belongs to Group 2.

The electronic configuration of strontium (Sr) = $[Kr]5s^2$

Hence, the correct answer is Strontium.

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Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

2 years ago

Which change is most likely to occur when a molecule of H2 and a molecule of I2 collide with proper orientation and sufficient e

bazaltina [42]

The collision of the molecules between the hydrogen molecule or H2, and an iodine molecule or I2, provided there would be a sufficient energy is that the system would eventually undergo a chemical change wherein a new chemical compound would be formed from these two molecules.

8 0

3 years ago

Read 2 more answers

Based on this chart, which of the following statements describes sources of

Vera_Pavlovna [14]

Answer:A is correct

Explanation:

4 0

2 years ago

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Write the balanced chemical equation for the following: The destruction of a marble statuary by acid rain: aqueous nitric acid r

Mashcka [7]

Answer: The coefficient in front of nitric acid is 2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas. Liquids are represented by (l) and gases are represented by (g).

The balanced chemical reaction is :

$CaCO_3(s)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+CO_2(g)+H_2O(l)$

7 0

2 years ago

The microscopes used today are just like the ones used by Leeuwenhoek and Hooke. True or false

Dmitry [639]

Answer:

False

Explanation:

While we do know that A. Leeuwenhoek used a simple microscope that consisted of only 1 lens, Hooke used a compound microscope. Although, after trying a compound microscope, Hooke found out that it strained his eyes and continued to use a simple microscope for his Micrographia.

Thus, we can say that the (compound) microscopes used today are different than the (simple) microscope used by Hooke and Leeuwenhoek.

6 0

2 years ago