Which expression can be used to calculate the density of a 129.5-gram sample of bronze that has a volume of 14.8 cubic centimete
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This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.
<h3>Balancing chemical equations:</h3>In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.
For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:
As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:
Learn more about balancing chemical equations: brainly.com/question/8062886
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!