Answer:
340 J·K⁻¹mol⁻¹
Explanation:
I looks like the standard entropy increases approximately linearly with the number of C atoms.
I plotted S° vs the number of C atoms and got the graph shown below.
It appears that S° for four carbon atoms should be about 340 J·K⁻¹mol⁻¹.·
Answer:
tungsten
Explanation:
I hope it's helpful for you
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:
0.000237mL
Explanation:
0.237 x 10^-6L = 0.000000237L = mL
0.000237mL
The specific heat capacity of liquid water is 4.186 J/gm K. This means that each gram of liquid water requires 4.186 Joules of heat energy to raise its temperature by one degree Kelvin. One molar mass of water is equivalent to 18 grams. Therefore, the molar heat capacity becomes the product of 4.186 and 18