Answer:
The compound is N2O4
Explanation:
We have certain important pieces of information about the compound;
1) it is an oxide (a binary compound of nitrogen and oxygen)
2) there are no N-N bonds present
3) there are no O-O bonds present
Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen
Relative atomic mass of oxygen = 16
Relative atomic mass of nitrogen = 14
We now deduce the empirical formula
Nitrogen. Oxygen
25.9/14. 74.1/16
1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)
1 2
Empirical formula is NO2
To find the molecular formula
(NO2)n = 108
(14+2(16))n= 108
46n=108
n= 108/46
n= 2
Therefore molecular formula= N2O4
Group I and Group II have the same number of outermost electrons as you go down each group but the shells increase therefore as you go downward it becomes much easier to remove electrons because of its wide radius however group 7 and 6 have seven and six electrons in their outermost shell respectively. Therefore down the group it is much easier to attract electrons and across the period it is much harder because the number of electrons on the outermost shell increase
Answer:
The correct answer is C.
Explanation:
To solve this problem, we should first recall possible units for mass and volume. Mass can be represented with units of grams, kilograms, etc. Volume has units of cubic meters, cubic centimeters, etc.
Therefore, since we know that density is mass divided by volume, the only answer that makes sense here is C. grams per cubic centimeter (where per signifies that grams are being divided by cubic centimeters).
Hope this helps!
Answer : The concentration of 
at equilibrium is 0 M.
Solution : Given,
Concentration of = 0.0200 M
Concentration of 
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of 
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)
By solving the term, we get:
Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of at equilibrium is 0 M.
A) a hypothesis hope it helped :)
