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icang [17]
3 years ago
9

Which bicylic compounds are unstable at room temperature?

Chemistry
2 answers:
jek_recluse [69]3 years ago
8 0

Answer:

the compounds are:

Chlorine Trifluoride (CLF3)

Substance N (another way to call CLF3)

Azido Azide Azide (C2N14)

Explanation:

The instability of the C2N14 is beyond our handling capabilities. Minor tests of load and friction led to explosive decomposition.

This is the most explosive compound known, ironically two nitrogen atoms linked with a triple covalent bond is the most stable molecule, but in the case of Azido Azide Azide none of its 14 nitrogen atoms is linked by a triple bond, which makes it very unstable.

Substance N in 1930 was a new compound was discovered by Ruff and Krug in Germany. It was too volatile, so it was ignored, until a few years after it sparked interest in Nazi scientists. They named the compound substance n and it showed very particular properties:

· Boils at room temperature and produces toxic gas

· If the gas is ignited it burns at more than 2,400 degrees Celsius

· Explodes on contact with water

· If combined with coal it forms an explosive that detonates on contact with anything else

Seeing these properties and that substance n was so good at setting fire to things that were not flammable like glass or sand the Germans decided to use it.

It is a colorless gas or a highly reactive white solid with a sweet, suffocating odor. It is transported as a greenish-yellow liquid.

It is used in rocket boosters and in the processing of fuels for atomic reactors.

astraxan [27]3 years ago
8 0

Answer:

Explanation:

solution is stated in the attached document

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PLEASE HELP ASAP!
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Answer:

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2 years ago
If 25.5g of sodium thiosulphate was dissolved in 40g of distilled water at 25°C,
zhenek [66]

Answer:

Formula: Na2S2O3

we get solubility.

Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .

Solution given:

mass of sodium thiosulphate [m1]=25.5g

mass of water [m2]=40g

at temperature [t]=25°C

we have

<u>solubility in g/dm^3</u> :\frac{solute in gram}{solvent in gram} *100

  • =\frac{25.5}{40}*100
  • =63.75g /litre=63.75g/dm³

<u>solubility in g/dm^3 :63.75g/dm³</u>

<u>n</u><u>o</u><u>w</u>

solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³

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