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Readme [11.4K]
3 years ago
7

"how many molecules of carbon dioxide exit your lungs when you exhale 5.00 × 10−2 mol of carbon dioxide, co2?"

Chemistry
2 answers:
IrinaVladis [17]3 years ago
5 0
We are going to use Avogadro's constant to calculate how many molecules of

carbons dioxide exist in lungs:

when 1 mole of CO2 has 6.02 x 10^23 molecules, so how many molecules in

CO2 when the number of moles is 5 x 10^-2

number of molecules = moles of CO2 * Avogadro's number

                                     = 5 x 10^-2  * 6.02 x 10^23

                                     = 3 x 10^22 molecules 

∴ There are 3 x 10^22 molecules in CO2 exist in lungs 
SpyIntel [72]3 years ago
3 0
The number of CO2  molecules  that   exit  is calculated   as   follows

by use  of Avogadro   law  constant

1   mole  = 6.02 x10^23  molecules
5.00 x10^-2 mol   =?   molecules

{( 6.02  x10^23)x (5.00  x10^-2) / 1}  = 3.01 x10^22  molecules

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3 0
2 years ago
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The elements in Groups 1 AU) and 7A(17) are all quite reactive. What is a major difference between them?
Yuliya22 [10]

Explanation:

Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium, and francium.

All these elements are metals and every element of this group has 1 valence electron. So, in order to attain stability they will readily lose their valence electron.

Hence, elements of group 1A are very reactive.

On the other hand, elements of group 7A are also known as halogen group. Elements of this group are fluorine, chlorine, bromine, iodine, and astatine.

All these elements are non-metals and every element of this group has 7 valence electrons. So, in order to completely fill their octet these elements gain 1 electron from a donor atom.

Therefore, these elements are alo reactive in nature.

But the major difference between elements of group 1A and group 7A is that elements of group 1A are metals but elements of group 7A are non-metals.    

8 0
4 years ago
A competitive inhibitor of an enzyme-catalyzed reaction - cannot bind to the active site. - always interferes with product relea
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Answer:

is usually structurally similar to the substrate.

Explanation:

Competitive inhibitors resemble normal substrate and binds to enzyme at the active site usually and prevents substrate from binding.

Active sites are main location for the substrate-enzyme binding. These sites involve weak as well as reversible bonds between the substrate and the enzyme. These inhibitors bind to the active sites and form weak and  reversible bonds. Competitive inhibitors can be dissociated from active site by increasing concentration of the substrates. Substrates has to compete for active site and displace the bound competitive inhibitors.

<u>Hence, correct option is - is usually structurally similar to the substrate.</u>

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3 years ago
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Answer: capitalized?

6 0
3 years ago
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Answer:

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In p-hydroxybenzoic acid, -OH and -COOH groups remain in para position.

The directing effect of -OH and -COOH group act synergistically to react with three equivalent of bromine.

In para position of -OH group, -COOH remains attached along with a Br atom in a same carbon atom. During aromatization of benzene group, this -COOH group decomposes to give carbon dioxide.

Reaction scheme has been shown below.

7 0
3 years ago
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