The total charge on an atom comes from protons and electrons.
The proton is positively charged while the electron is negatively charged. A neutral atom would have an equal number of protons and electrons.
An atom with more protons than electrons will be positively charged while those with more electrons than protons will be negatively charged.
More on the atom can be found here: brainly.com/question/1641336
If 4 moles of P is used by 5 mole of O2
then....0.489 moles will be used by 5/4 × .489 = .611 moles of O2
so .611 moles
so if 4 moles of P is burnt , 1 mole of P4O10 is produced ....so for .489 moles...... .489/4=.122 moles !
so mass will be .122× 283.89 = 34.7 grams
so first ans is .611 moles and second is 34.7 grams !
if you have any problem regarding this , just comment !!!
There are 3 equations involved in manufacturing Nitric Acid from Ammonia.
First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O
Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4
Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO
Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)
Convert the exact number back into tons, and your answer is: 419.332775 tons.
Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)
Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n= 
= 
= density
P = 
density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 = 
M = 34.61
to calculate the Kc
Kc=![\frac{ [NO] [O2]}{NO2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BNO%5D%20%5BO2%5D%7D%7BNO2%7D)
x M NO2 + 
M NO+ 
M O2
Putting the values as molecular weight of NO2, NO,O2
34.61= 
x= 0.33
Kc= 
putting the values in the above equation
Kc = 0.00662
6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.
0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.
