Can anyone please help me with my Chemistry homework
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Can anyone please help me with my Chemistry homework

[From: ] [author: ] [Date: 11-07-04] [Hit: ]
Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass = 14.00 amu ) and nitrogen-15 (mass = 15.How many NBr3 molecules of different masses can exist?00amu ). Bromine also has two naturally occurring isotopes: bromine-79 (mass = 78.92 amu ) and bromine-81 (mass = 80.......
1. Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass = 14.00 amu ) and nitrogen-15 (mass = 15.How many NBr3 molecules of different masses can exist?00amu ). Bromine also has two naturally occurring isotopes: bromine-79 (mass = 78.92 amu ) and bromine-81 (mass = 80.92 amu).

How many NBr3 molecules of different masses can exist?

Determine the mass (in amu ) of each of them.

2.

Symbol Z A Number of Protons Number of Electrons Number of Neutrons Charge

-- 25 -- -- 13 2+
-- 22 48 -- 18 -- --
-- 16 -- -- -- 16 2 -
-- 71 -- -- -- --
-- -- -- 82 80 125 --

a) Fill in the blanks in Symbol column of the table.
Express your answers as ions.

b)Fill in the blanks in A column of the table.

c)Fill in the blanks in Number of Protons column of the table.

d)Fill in the blanks in Number of Electrons column of the table.

e)Fill in the blanks in Number of Neutrons column of the table.

f)Fill in the blanks in Charge column of the table.


please explain step by step

-
*** 1 ***
think permutations..

N.. B1.. B2.. B3
14..79.. 79.. .79
14..79.. 79.. .81
14..79.. 81.. .81
14..81.. 81.. .81
15..79.. 79.. .79
15..79.. 79.. .81
15..79.. 81.. .81
15..81.. 81.. .81

now sum up the masses for those 8 possibilities..
14.00 + 78.92 + 78.92 + 78.92 =
14.00 + 78.92 + 78.92 + 80.92 =
etc

*** 2 ***

I'm having a hard time with your table..if I read it like this

Sym.. Z.. #p... #e... #n.. .charge
.. .. .. 25.. .. .. .... ... 13.. .2+
.. .. .. 22.. 48.. .... .. 18
.. .. .. 16.. .. .. .... ... 16.. .2-
.. ... .. 71.. ... .. .. .. .. . . .. .
.. ... .. .. ...82.. .80.. .125

clearly rows 2 and 3 are impossible and row 4 is not solveable.

#p = atomic#.. once you have #p.. look up the atomic # and symbol on a periodic table
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