For this problem, we should use the Henry's Law formula which is written below:
P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration
Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater
Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>
Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g
Answer: 51.45 grams of excess reagent is left after the completion of reaction.
Explanation: For the calculation of moles, we use the formula:
....(1)
Given mass = 92 grams
Molar mass = 28g/mol
Putting values in equation 1, we get:
![Moles=\frac{92g}{28g/mol}=3.285moles](https://tex.z-dn.net/?f=Moles%3D%5Cfrac%7B92g%7D%7B28g%2Fmol%7D%3D3.285moles)
- For
![Cr_2O_3](https://tex.z-dn.net/?f=Cr_2O_3)
Given mass = 112 grams
Molar mass = 116g/mol
Putting values in equation 1, we get:
![Moles=\frac{112g}{116g/mol}=0.965moles](https://tex.z-dn.net/?f=Moles%3D%5Cfrac%7B112g%7D%7B116g%2Fmol%7D%3D0.965moles)
The reaction follows:
![2Cr_2O_3(s)+3Si(s)\rightarrow 4Cr(l)+3SiO_2(s)](https://tex.z-dn.net/?f=2Cr_2O_3%28s%29%2B3Si%28s%29%5Crightarrow%204Cr%28l%29%2B3SiO_2%28s%29)
By Stoichiometry,
2 moles of
reacts with 3 moles of silicon
So, 0.965 moles of
reacts with =
= 1.4475 moles of Silicon.
As, the moles of silicon is more than the required amount and is present in excess.
So, the excess reagent for the reaction is Silicon.
Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles
To calculate the amount of Silicon left in excess is calculated by using equation 1:
![1.8375=\frac{\text{Given mass}}{28}](https://tex.z-dn.net/?f=1.8375%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B28%7D)
Amount of Silicon in excess will be 51.45 grams.
39 % = 950 mg = 0.75 g
<span>100 % = ? </span>
<span>? = 100 / 39 * 0.95 = 2.44g or 2.4358974 of NaCl</span>
Answer:
I don't know
Explanation:
I'm sorry I didn't learn this in my life,I have never reach in this level
To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>