1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4
Answer:
The plane with aluminium can lift more mass of passangers than the plane of steel.
Explanation:
The total mass the airplane canc lift is:

For aluminium:


and
![V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]](https://tex.z-dn.net/?f=V_%7Bfuselage%7D%3D%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D)
where:
- L is lenght
- D is diameter
- e is thickness
![m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BAl%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Al%7D)
For steel (same procedure):
![m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BSteel%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Steel)
Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.
Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.
I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.
First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,
For
Benzyl chloride,

=

Solving for X,
X =

X = 78.79 mg
For PPh₃:

=

Solving for X,
X =

X = 163.27 mg
Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,
when

=

Solving for X,
X =

= 231.57 mg
Now, calculate reactants mass with respect to 231.57 mg
when

=

Solving for ,
X =

=
82.93 mg of Benzyl chloride
when

=

Solving for ,
X =

=
171.85 mg of PPh3
So, reaction was started with reacting
82.93 mg of Benzyl Chloride and
171.85 mg of Triphenyl Phosphine.
Answer:
Mole fraction of nitrogen = 0.52
Explanation:
Given data:
Temperature = 31.2 °C
Pressure = 870.2 mmHg
Volume = 15.1 L
Mass of mixture = 24.1 g
Mole fraction of nitrogen = ?
Solution:
Pressure conversion:
870.2 /760 = 1.12 atm
Temperature conversion:
31.2 + 273 = 304.2 K
Total number of moles:
PV = nRT
n = PV/RT
n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K
n = 16.9 L.atm. /25 L.atm. mol⁻¹
n = 0.676 mol
Number of moles of nitrogen are = x
Then the number of moles of CO₂ = 0.676 - x
Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)
24.1 = 28x + ( 29.7 -44x)
24.1 - 29.7 = 28x - 44x
-5.6 = -16 x
x = 0.35
Mole fraction of nitrogen:
Mole fraction of nitrogen = moles of nitrogen / total number of moles
Mole fraction of nitrogen = 0.35 mol / 0.676 mol
Mole fraction of nitrogen = 0.52