Explanation:
The given data is as follows.
= 30.0 sec, 
= 5 min = 
= 300 sec
= 12.0 min = 
= 720 sec
Formula for adjusted retention time is as follows.
= 300 sec - 30.0 sec
= 270 sec
= 720 sec - 30 sec
= 690 sec
Formula for relative retention (
) is as follows.
= 
= 2.56
Thus, we can conclude that the relative retention is 2.56.
Your compound is
.
Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.
Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³
Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
=(346/101.068)*(1/0.75)
=4.56 mol dm⁻³
Answer:
The answer to your question is 50 moles of O₂
Explanation:
Balanced Chemical reactions
1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)
moles of N₂(g) = 20 moles
moles of O₂(g) = ?
Process
1.- Calculate the moles of NH₃
1 mol of N₂ ------------- 2 moles of NH₃
20 moles of N₂ --------- x
x = (20 x 2) / 1
x = 40 moles of NH₃
2.- Calculate the moles of O₂
4 moles of NH₃ -------------- 5 O₂
40 moles of NH₃ ------------ x
x = (40 x 5) / 4
x = 200 / 4
x = 50 moles of O₂
Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
