Answer: 1741289L
Explanation:
1 gallon = 3.78541 L
4.6×10^5 gallons = 4.6×10^5 x 3.78541 = 1741289L
Answer:
<u>STEP I</u>
This is the balanced equation for the given reaction:-
<u>STEP II</u>
The compounds marked with (aq) are soluble ionic compounds. They must be
broken into their respective ions.
see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).
On breaking them into their respective ions :-
- 2KOH -> 2K+ + 2OH-
- H2SO4 -> 2H+ + (SO4)2-
- K2SO4 -> 2K+ + (SO4)2-
<u>STEP III</u>
Rewriting these in the form of equation
<u>STEP </u><u>IV</u>
Canceling spectator ions, the ions that appear the same on either side of the equation
<em>(note: in the above step the ions in bold have gotten canceled.)</em>
This is the net ionic equation.
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- KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.
[Alkali metal hydroxides are the only halides soluble in water ]
You need the set of reactions that goes from ammonia to nitric acid.
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1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
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Answer: i need that answer to
Explanation: