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Ludmilka [50]
3 years ago
15

Why are atoms considered the smallest particles of an element?

Chemistry
2 answers:
77julia77 [94]3 years ago
3 0
Atoms are considered the smallest particles of an element because they are the basic unit. They cannot be broken down into any smaller substance. 
Vlada [557]3 years ago
3 0
They are considered the smallest part of an element because when you break down an element into smaller and smaller parts, theoretically when you reach thed smallest part and you cant break it any more, it will be one atom of whatever element you are breaking down.
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Dovator [93]

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

3 0
3 years ago
How many grams of carbon are in 0.24 moles of carbon?
Ronch [10]

Answer:

I'm converting this if I could remember how

2.882568

2 110321/ 125000

T-T sorry if I'm wrong I have bad memory

so I recommend not using my answer at all,

if that is even how y'all write it.

4 0
3 years ago
Please help me with letter a question
Colt1911 [192]
Rise, decrease, away from ocean, towards land
7 0
3 years ago
What is one pro and one con of mining?
blondinia [14]

Answer:

Pro exercise con suffication

Explanation:

...

8 0
3 years ago
There is a 30g of be-11 it has a half-life of about 14 seconds how much will be left in 28 seconds
Lelu [443]

There will be 7.5 g of Be-11 remaining after 28 s.

If 14 s = 1 half-life, 28 s = 2 half-lives.

After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.

After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.

In symbols,

<em>N</em> = <em>N</em>₀(½)^<em>n</em>

where

<em>n</em> = the number of half-lives

<em>N</em>₀ = the original amount

<em>N</em> = the amount remaining after <em>n</em> half-lives

6 0
3 years ago
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