How many iron atoms are in Fe(NO3)2? A. 6 B. 1 C. 3 D. 2

olya-2409 [2.1K]

The answer is head-to-tail joining of monomers. Monomer used in condensation have two functional groups that combine to form amide and ester linkages. When this reaction occurs, water molecules is removed and that is why it is called a condensation reaction.

8 0

3 years ago

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How is hydrogen different from alkali metals?

Irina18 [472]

Hydrogen is different from alkali metals because it also exhibits the properties of inert gas

5 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

2 years ago

Nuclear decay 27Al + He 》 39P

dybincka [34]

Answer:

$_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

Explanation:

The unbalanced nuclear equation is

$^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}$

We can insert the subscripts, because these are the atomic numbers of the elements

$_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

$_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}$

6 0

3 years ago

A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t

lidiya [134]

Answer: Copper (I) iodide will precipitate first.

Explanation:

We are given:

$K_{sp}$ of CuCl = $1.0\times 10^{-6}$

$K_{sp}$ of CuI = $5.1\times 10^{-12}$

Concentration of $Cl^-\text{ ion}=0.021M$

Concentration of $I^-\text{ ion}=0.017M$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

For CuCl:

$K_{sp}=[Cu^+][Cl^-]$

Putting values in above equation, we get:

$1.0\times 10^{-6}=[Cu^+]\times 0.021$

$[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M$

Concentration of copper (I) ion = $4.76\times 10^{-5}M$

For CuI:

$K_{sp}=[Cu^+][I^-]$

Putting values in above equation, we get:

$5.1\times 10^{-12}=[Cu^+]\times 0.017$

$[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M$

Concentration of copper (I) ion = $3.00\times 10^{-10}M$

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0

3 years ago