Your answer to 2.5*22.56 is 56.25
Answer & Explanation:
Los electrones externos se encuentran más lejos del núcleo. El número de electrones en la capa más externa (electrones de valencia) de un átomo en particular determina su reactividad (tendencia) a formar enlaces químicos con otros átomos.
Los electrones internos son los más cercanos al núcleo. Protegen los electrones de valencia del núcleo, reduciendo la carga nuclear efectiva.
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
Stationary Front: a front that is not moving. When a warm or cold front stops moving, it becomes a stationary front.
Answer : The enthalpy of the given reaction will be, -1048.6 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The main reaction is:
The intermediate balanced chemical reactions are:
(1) 
(2) 
(3) 
(4) 
(5) 
Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :
(1) 
(2) 
(3) 
(4)
(5) 
The expression for enthalpy of main reaction will be:
Therefore, the enthalpy of the given reaction will be, -1048.6 kJ
