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Answer:
%C = 56,1%
%H = 5,5%
%Cl = 27,6%
%N = 10,8%
Explanation:
The moles of CO₂ are the same than moles of C in the herbicide.
Moles of H₂O are ¹/₂ of moles of H in the herbicide.
Moles of CO₂ are obtained using:
n = PV/RT
Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K
moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>
In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>
As you have 55,14 mg of Cl, the mg of N are:
200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N
Thus, precent composition of the herbicide is:
%C = ×100 = 56,1%C
%H = ×100 = 5,5%H
%Cl = ×100 = 27,6%Cl
%N = ×100 = 10,8%N
I hope it helps!
Answer:
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 1.25 atm
V = 25.0 L
n = 2.10 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
1. Temperature in kelvins
2. Temperature in degrees Celsius
Answer:
<h2>36.09 L</h2>Explanation:
The initial volume can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume.
Since we're finding the initial volume
We have
We have the final answer as
<h3>36.09 L</h3>Hope this helps you