1 mol N₂ - 2 mol NH₃
6 mol N₂ - x mol NH₃
x=2×6/1=12 mol
12 mol NH₃
Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.
T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
Answer:
1.2 × 10^22 atoms.
Explanation:
Firstly, cations refers to the positively charged atom in the ionic compound, which is Na+.
Given the mass of NaCl as 1.17g, the number of moles of NaCl can be calculated this:
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
Mole = mass/molar mass
Mole = 1.17/58.5
Mole = 0.02moles
Using Avagadro's number, 6.022 × 10^23 atoms of Na+ are in 1 mole of NaCl.
In 0.02 moles of NaCl, there are 0.02 × 6.022 × 10^23 of Na+
0.1204 × 10^23 atoms
1.2 × 10^22 atoms of Na+ (cation)
Al, Si, Mg
All other elements seem more necessary in life
Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: 
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29)
; Silver hydroxide at higher temperatures decomposes into black silver oxide: ![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29)
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate: 
. The latter precipitate is yellow.
