1 mol of any particles has 6.02 * 10 ²³ particles.
If we look at 1 NH3 (1 mol NH3 or 1 molecule NH3), we can see that 1 molecule NH3 has 1 atom of N and 3 atoms of H; also 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms.
So, 1 mol of NH3 has 1 mol of N atoms,
and 2.79 mol NH3 have 2.79 mol of N atoms.
2.79 mol of N atoms* 6.02 * 10 ²³ N atoms/ 1 mol N atoms = 1.68*10²⁴ N-atoms
Answer is 1.68*10²⁴ N-atoms.
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
P = 14.9258934 atm
N2 + 3H2 --> 2NH3
Answer: 6 moles of hydrogen are needed to react with two moles of nitrogen.
Explanation:
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Answer:
Here's what I get
Explanation:
A plant extract is a mixture because it contains different substances: acetone or ethanol, chlorophylls A and B, carotene and xanthophylls.
It is homogeneous because it is a solution. There is only one phase: the liquid phase. You cannot see the pigments as separate phases.
You can separate the pigments by paper, thin layer, or column chromatography.
Many schools use paper chromatography, because paper is cheap.
As the mixture of pigments follows the solvent up the paper, they separate into different coloured bands according to their attractive forces to the cellulose in the paper.
The chlorophylls are strongly attracted to the paper, so they don't travel very far.
The nonpolar carotene molecules have little attraction to the polar cellulose, so they are carried along by the solvent front.
