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Nesterboy [21]
4 years ago
8

Name the compound Fe(NO2)2?

Chemistry
2 answers:
Contact [7]4 years ago
5 0
The compound Fe(NO2)2 is Iron (II) Nitrite. 
Hope I could help :) 
kotykmax [81]4 years ago
4 0
The compounds name is iron nitrite
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Which metal ions can be precipitated out of solution as chlorides?a. Ag+, Hg2+, Co2+b. Cu2+, Cd2+, Bi3+ c. Ag+, Hg22+, Pb2+ d. N
Fudgin [204]

Answer:

Explanation:

In the qualitative analysis of metal salts , we see that in group I , metal chlorides are precipitated out . It is so because their metal chlorides are insoluble in water  .

In this group following metal ions are present

Ag+,

Hg₂²⁺

Pb²⁺

8 0
3 years ago
Explain why the benzene molecule usually reacts with electrophiles<br>​
Genrish500 [490]

Answer:

Explanation:Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules.

3 0
3 years ago
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What is the pH of a solution whose H + concentration is 4.0 10 –9?
Dmitrij [34]

Answer:

he pH of a solution is defined as the negative log10 [H+] ... 1 x 10-11. 11. Acidic Solution. 1 x 10-4. 4. 1 x 10-10. 10. 1 x 10-5. 5. 1 x 10-9.

Explanation:

m

5 0
3 years ago
The rate constant for this second‑order reaction is 0.190 M − 1 ⋅ s − 1 0.190 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product
Mamont248 [21]

Answer:

9.1 seconds

Explanation:

Given that for a second order reaction

1/[A]t = kt + 1/[A]o

Where [A]t= concentration at time = t= 0.340M

[A]o= initial concentration = 0.820M

k= rate constant for the reaction=0.190m-1s-1

t= time taken for the reaction (the unknown)

Hence;

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

5 0
3 years ago
Based on the trend in the table, hypothesize what the volume of the sample at 400K would be.
mariarad [96]

Answer: The volume of sample at 400 K is 285cm^3

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T     (At constant pressure and number of moles)

{\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1 = initial volume of gas  = 257cm^3

V_2 = final volume of gas  = ?

T_1 = initial temperature of gas  = 360 K

T_2 = final temperature of gas  = 400 K

Putting in the values we get:

{\frac{257}{360}=\frac{V_2}{400}

V_2=285cm^3

Thus volume of sample at 400 K is 285cm^3

6 0
3 years ago
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