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Jlenok [28]
3 years ago
11

Question 7 (1 point)

Chemistry
1 answer:
Anastasy [175]3 years ago
5 0

Answer:

2.28bar

Explanation: Boyle's law P1V1=P2V2 manipulate formula in favor of V2 the new formula should beP1V1/V2

0.895*318/125=2.2768 but to same sigfig it is 2.28

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Because the nuclear charge increases across a period and so it has a stronger pull on the outer electrons and will pull in the radius
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Organisms such as grass and lichens are classified as
mash [69]

I think the answer is producers

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3 0
3 years ago
Since the number of atoms in a substance is so large , a unit to count them was created. This unit is the number of atoms in 12
larisa86 [58]

The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.


(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.

7 0
3 years ago
This is a test please help
GenaCL600 [577]

Answer:

remaining still during the night.

Explanation:

3 0
3 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
3 years ago
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