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:
-3
Explanation:
The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The complex anion here is [Cr(CN)6]3-.
Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),
so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.
.
Don’t eat or drink in labs
Dress for the lab; don’t wear open toed shoes
Dispose of lab waste properly
No horse play
Don’t taste or sniff things in the lab
Tire your hair back
It is called a waxxing gibbous, pls brainliest
The answer is “D Tissue”
Explanation:
when you workout you tear down your muscles (tissue) and after your done working out your body rebuilds tissue
hope this helps
