NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
Answer:
2.94 x 
Explanation:
First we need to find out how many moles of ammonia there are, using the formula: Mass = mr x moles.
We know the mass is 83.1g, now we need to find the mR of ammonia - NH3.
N = 14, H = 1, so 14 + (3x1) = an mr of 17.
Moles = mass/ mr = 83.1/17 = 4.8882
Now we can multiply the moles by avogadro's constant to find the number of molecules:
4.8882 x (6.02 x 
) = 2.94 x molecules of ammonia
Answer:
True
Explanation:
According to Aufbau's principle "sublevels with lower energies are filled up before those with higher energies".
Sublevels do not fill up in numerical order but there is a certain manner in which they are filled. The pattern is shown below:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p e.t.c
We see that the 4s gets filled before the 3d sublevel.
