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Brilliant_brown [7]
3 years ago
14

Is flammability a form of chemical change or physical change

Chemistry
2 answers:
Vinvika [58]3 years ago
6 0
Chemical change C===3
Gnoma [55]3 years ago
3 0

Answer:

When a fuel burns, it combines with oxygen int the air and changes into the substances water and carbon dioxide.

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saul85 [17]
Tell your teacher. They'll know what to do and it's best to report it to them.
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Substances that are likely to dissociate it water
Neporo4naja [7]

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sodium bromide (NaBr) potassium hydroxide (KOH) magnesium chloride (MgCl2) silicon dioxide (SiO2) sodium oxide (Na2O)

Explanation:

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4 years ago
1 cholo 2 floro 4 nitro + Na^+​
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Answer:

Incomplete Q.

GOOD LUCK FOR THE FUTURE! :)

4 0
3 years ago
Read 2 more answers
What are the isomers of butanol​
Juliette [100K]

There are three other structural isomers of 1-butanol: 2-butanol (sec-butyl alcohol), 2-methyl-1-propanol (isobutyl alcohol), and 2-methyl-2-propanol (tert-butyl alcohol). 2-Butanol, or sec-butanol, or sec-butyl alcohol, or s-butyl alcohol, is a four-carbon chain, with the OH group on the second carbon.

Chemicals of this type: Ethanol

Hope this helps

7 0
3 years ago
50 ml of 0.2 M acetic acid is shaken with 10 g activated charcoal. The concentration of acetic acid is reduced to 0.5 times the
Olegator [25]

Answer:

0.03 g_{acid}/g_{charcoal}

Explanation:

The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.

So, we need to calculate the mass of the acetic acid as follows:

m = n*M = C*V*M

Where:

n: is the number of moles = C*V

M: is the molecular mass =  60.052 g/mol

C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L

V: is the volume = 50 ml = 0.050 L

m_{acid} = C*V*M = 0.10 mol/L*0.050 L*60.052 g/mol = 0.30 g

Now, the amount of solute adsorbed per gram of the adsorbent is:

\frac{m_{acid}}{m_{charcoal}} = \frac{0.30 g}{10 g} = 0.03 g_{acid}/g_{charcoal}

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.

I hope it helps you!

3 0
3 years ago
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