Answer:
Would it be<em><u> 7.69 seconds</u></em>?
Explanation:
Every balanced chemical reaction demonstrates the law of conservation of mass where the reactants have the same number of atoms as the products. For example, the production of ammonia gas.
N2 (g) + 3H2 (g) -> 2NH3 (g) you can see that there are two nitrogen atoms and six hydrogen atoms on both sides of the equation.
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:
the 4s subshell is lower in energy than the 3d subshell
Explanation:
In the element number 24 which is chromium
the electronic configuration should be according to Aufbau principle
will be = ![1s^22s^22p^63s^23p^64s^23d^4](https://tex.z-dn.net/?f=1s%5E22s%5E22p%5E63s%5E23p%5E64s%5E23d%5E4)
but in reality it is = ![1s^22s^22p^63s^23p^64s^13d^5](https://tex.z-dn.net/?f=1s%5E22s%5E22p%5E63s%5E23p%5E64s%5E13d%5E5)
The 6 electrons in the outermost shell will be divided as 5 in 3d subshell and 1 in 4s shell, because half filled orbital
will be more stable than
.