Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
Answer:
Q was < K. Partial pressure of hydrogen decreased, iodine increased
Explanation:
<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>
Based on the equilibrium:
H2(g) + I2(g) ⇄ 2HI(g)
K of equilibrium is:
K = [HI]² / [H2] [I2]
<em>Where [] are concentrations at equilibrium</em>
And Q is:
Q = [HI]² / [H2] [I2]
<em>Where [] are actual concentrations of the reactants.</em>
<em />
When the reaction is in equilibrium, K=Q.
But as [I2] is increased, Q decreases and Q was < K
The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased
Hi there!
• Avogadro's number = 6.023 × 10²³
• No.of molecules in N = 1.806 × 10²² [ Given ]
It's known that :-
Number of molecules = Moles × Avogadro's number
=> 1.806 × 10²² = Mol. × 6.023 × 10²³
=> Mole =

=> Moles = 0.03 mol.
Hence, 0.03 mol. is th' required answer.
~ Hope it helps!
The answer is: Cl2.
Chlorine is diatomic molecule made of two chlorine atoms.
Diatomic molecules are molecules made of two atoms.
They can be homonuclear (molecule made of two atoms of the same element) and heteronuclear (molecule made of two different atoms).
Chlorine (Cl) is halogen element.
Halogen elements are in group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). They are very reactive and easily form many compounds.