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Greeley [361]
3 years ago
11

Name diatonic elements

Chemistry
2 answers:
Volgvan3 years ago
8 0
I assume you mean diatomic elements.
Those are hydrogen, oxygen, nitrogen, fluorine, chlorine, bromine, and iodine. They all exist in the form X_{2} (H_{2},  O_{2} ,  N_{2} ,  F_{2} ,  Cl_{2} ,  Br_{2} ,  I_{2}). Where they are all bonded to another one of themselves (a hydrogen atom binds to another hydrogen atom, a fluorine atom binds to another fluorine atom, etc.)
Tresset [83]3 years ago
3 0
Hydrogen, oxygen, nirtogren, fluorine, chlorine, bromine, iodine
You might be interested in
Help me out please!!
malfutka [58]

Answer:6.48*10^{-2}

Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

please remember this number and commit it to your memory: avogadros number

6.02 * 10^{23}

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is 6.02 * 10^{23} it is confusing at first but hopefully this helps you to understand.

now that we understand this..... lets perform this calculation with a calculator

(3.90 * 10^{22}) / (6.02 * 10^{23})

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten.....    22 minus 23 equals -1

so we have 0.6478 * 10^{-1}

now this negative 1 thing is annoying so lets just make it to the power of 0

0.06478 * 10^{0}

and anything to the power of 0 just becomes 1.

0.06478

so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648

put this into scientific notation we get 6.48*10^{-2}

5 0
3 years ago
2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
3 years ago
What are 4 ways that water can go through a physical change?
zlopas [31]
Being frozen, staying a liquid, becoming ice, and becoming a gas (steam)
5 0
3 years ago
Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their
dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Pure Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Br and Br,

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00         (Non Polar/Pure Covalent)

 

For N and O,

                    E.N of Oxygen      =   3.44

                    E.N of Nitrogen     =   3.04

                                                   ________

                    E.N Difference             0.40           (Non Polar/Pure Covalent)

 

For P and H,

                    E.N of Hydrogen       =   2.20

                    E.N of Phosphorous  =   2.19

                                                              ________

                    E.N Difference                  0.01          (Non Polar/Pure Covalent)

 

For K and O,

                    E.N of Oxygen          =   3.44

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                2.62              (Ionic)

6 0
3 years ago
Trey was asked to build a working model of a landform for a science project. He chose to build a model of a volcano.
maksim [4K]
Can you please give me more details so I can help you
3 0
3 years ago
Read 2 more answers
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