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Galina-37 [17]
3 years ago
14

reativity of alkali metals increases down the group while reativity decreases down the group in helogens ​

Chemistry
1 answer:
BigorU [14]3 years ago
4 0

Complete question is;

Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons

Answer:

Explained below

Explanation:

Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.

Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.

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HHHAAAAALLLPPPP<br><br> It’s due tomorrow
iris [78.8K]

Answer:

D.

Explanation:

Only 0.0035% of the electromagnetic spectrum is visible to the human eye

Hopefully this helped :)

7 0
2 years ago
How many moles of oxygen are needed to react with 87 grams of aluminum
labwork [276]

Answer:

2.4 moles of oxygen are needed to react with 87 g of aluminium.

Explanation:

Chemical equation:

4Al(s)  + 3O₂(l)   → 2AlO₃(s)

Given data:

Mass of aluminium = 87 g

Moles of oxygen needed = ?

Solution:

Moles of aluminium:

Number of moles of aluminium= Mass/ molar mass

Number of moles of aluminium= 87 g/ 27 g/mol

Number of moles of aluminium= 3.2 mol

Now we will compare the moles of aluminium with oxygen.

                              Al         :         O₂

                               4          :         3

                               3.2       :         3/4×3.2 = 2.4 mol

2.4 moles of oxygen are needed to react with 87 g of aluminium.

5 0
2 years ago
Find the number of Grams
zaharov [31]

Answer: 462 g

Explanation:molar mass is M= 63.55 +2·(12.01+14.01)= 115.59 g/mol.

Mass m= n·M = 4.0 mol·115.59 g/mol= 462.36 g

8 0
2 years ago
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
3 years ago
12. If a solution has a pH of 3, what would the pOH value of that solution be? *
butalik [34]

Answer:

The correct answer is pOH= 11

Explanation:

From the aqueous acid-base equilibrium we know that

pH + pOH = 14

If we know pH, we can calculate pOH as follows:

pOH = 14 - pH

In this problem, the solution has a pH of 3, so:

pOH = 14 - 3 = 11

8 0
2 years ago
Read 2 more answers
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