Answer:
Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.
Explanation:
Percentage of element in compound :
(a) Urea, 
Molar mass of urea = 60 g/mol
Atomic mass of nitrogen = 14 g/mol
Number of nitrogen atoms = 2
(b) Ammonium nitrate, 
Molar mass of ammonium nitrate = 80 g/mol
Atomic mass of nitrogen = 14 g/mol
Number of nitrogen atoms = 2
(c) Nitric oxide, NO
Molar mass of nitric oxide = 30 g/mol
Atomic mass of nitrogen = 14 g/mol
Number of nitrogen atoms = 1
(d) Ammonia, 
Molar mass of ammona = 17 g/mol
Atomic mass of nitrogen = 14 g/mol
Number of nitrogen atoms = 1
Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.
Answer:
Below
Explanation:
Balanced form;
1.Benzene + Dioxygen = Carbon Dioxide + Water
2.Tricalcium phosphate +Carbon = Calcium phosphide + carbon monoxide
3.Nitrous acid react with oxygen to produce nitric acid.
4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.
5.Potassium react with bromine to produce potassium bromide
6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.
Water is the BL base if it accepted a proton from NH4.
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Answer:
3. Analizando la siguiente reacción química de formación del agua:
2 H2 (g) + O2 (g) 2 H2O (l)
Si se hacen reaccionar 10 g de hidrógeno con 10 g de oxígeno:
a) Plantee las razones molares de dicha ecuación
b) Determine si se consumen la totalidad de dichos reactivos. Demuéstrelo con los cálculos estequiométricos correspondientes.
Answer : The rate constant at 525 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 
R = gas constant = 8.314 J/mole.K
= initial temperature = 701 K
= final temperature = 525 K
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B2.57M%5E%7B-1%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B1.5%5Ctimes%2010%5E5J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B701K%7D-%5Cfrac%7B1%7D%7B525K%7D%5D)
Therefore, the rate constant at 525 K is,
