No,Biomass is organic material which has stored sunlight in the form of chemical energy. Biomass fuels include wood, wood waste, straw, manure, sugar cane, and many other byproducts from a variety of agricultural processes. Biomass is a renewable energy source because the energy it contains comes from the sun<span>.</span>
Mg reducing agent and Cu oxidising agent.
The question is incomplete, the complete question is:
Write the net ionic equation for the below chemical reaction:
(c): ![FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI](https://tex.z-dn.net/?f=FeCl_3%2B3NH_4OH%5Crightarrow%20Fe%28OH%29_3%2B3NH_4CI)
<u>Answer:</u> The net ionic equation is ![Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3OH%5E%7B-%7D%28aq%29%5Crightarrow%20Fe%28OH%29_3%28s%29)
<u>Explanation:</u>
Net ionic equation is defined as the equations in which spectator ions are not included.
Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.
(c):
The balanced molecular equation is:
![FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)](https://tex.z-dn.net/?f=FeCl_3%28aq%29%2B3NH_4OH%28aq%29%5Crightarrow%20Fe%28OH%29_3%28s%29%2B3NH_4Cl%28aq%29)
The complete ionic equation follows:
![Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3Cl%5E-%28aq%29%2B3NH_4%5E%2B%28aq%29%2B3OH%5E%7B-%7D%28aq%29%5Crightarrow%20Fe%28OH%29_3%28s%29%2B3NH_4%5E%2B%28aq%29%2B3Cl%5E-%28aq%29)
As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
![Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3OH%5E%7B-%7D%28aq%29%5Crightarrow%20Fe%28OH%29_3%28s%29)
Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH = ![\frac{15.0 g}{40 g/mol}=0.375 mol](https://tex.z-dn.net/?f=%5Cfrac%7B15.0%20g%7D%7B40%20g%2Fmol%7D%3D0.375%20mol)
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
![Molarity=\frac{Moles}{Volume(L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%7D%7BVolume%28L%29%7D)
![n=0.250 M\times 0.150 L=0.0375 mol](https://tex.z-dn.net/?f=n%3D0.250%20M%5Ctimes%200.150%20L%3D0.0375%20mol)
![NaOH+HNO_3\rightarrow NaNO_3+H_2O](https://tex.z-dn.net/?f=NaOH%2BHNO_3%5Crightarrow%20NaNO_3%2BH_2O)
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
![NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)](https://tex.z-dn.net/?f=NaOH%28aq%29%5Crightarrow%20Na%5E%2B%28aq%29%2BOH%5E-%28aq%29)
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
![=\frac{0.3375 mol}{0.150 L}=2.25 M](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.3375%20mol%7D%7B0.150%20L%7D%3D2.25%20M)
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Answer:
![P_{He}=9.06torr](https://tex.z-dn.net/?f=P_%7BHe%7D%3D9.06torr)
Explanation:
Hello there!
In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:
![P_{He}=x_{He}P_T](https://tex.z-dn.net/?f=P_%7BHe%7D%3Dx_%7BHe%7DP_T)
Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:
![n_{He}=8.00g\frac{1mol}{4.00g}=2.00mol\\\\ x_{He}=\frac{n_{He}}{n_{He}+n_{Ar}} \\\\ x_{He}=\frac{2.00mol}{2.00mol+8.60mol}\\\\x_{He}=0.189](https://tex.z-dn.net/?f=n_%7BHe%7D%3D8.00g%5Cfrac%7B1mol%7D%7B4.00g%7D%3D2.00mol%5C%5C%5C%5C%20x_%7BHe%7D%3D%5Cfrac%7Bn_%7BHe%7D%7D%7Bn_%7BHe%7D%2Bn_%7BAr%7D%7D%20%5C%5C%5C%5C%20x_%7BHe%7D%3D%5Cfrac%7B2.00mol%7D%7B2.00mol%2B8.60mol%7D%5C%5C%5C%5Cx_%7BHe%7D%3D0.189)
Then, we calculate the partial pressure as shown below:
![P_{He}=0.189 *48.0torr\\\\P_{He}=9.06torr](https://tex.z-dn.net/?f=P_%7BHe%7D%3D0.189%20%2A48.0torr%5C%5C%5C%5CP_%7BHe%7D%3D9.06torr)
Best regards!