<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO
<span>The two factors that determine whether or not a molecule is polar are if the individual bonds are even and the </span>shape<span> of the molecule. If the molecule is perfectly symmetric, the molecule will not be polar even if there are polar bonds present.</span>
Answer: Rate law=![k[A]^1[B]^2](https://tex.z-dn.net/?f=k%5BA%5D%5E1%5BB%5D%5E2)
, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 
Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![Rate=k[A]^x[B]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Ex%5BB%5D%5Ey)
k= rate constant
x = order with respect to A
y = order with respect to A
n = x+y = Total order
a) From trial 1: ![1.2\times 10^{-2}=k[0.10]^x[0.20]^y](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5Ex%5B0.20%5D%5Ey)
(1)
From trial 2: ![4.8\times 10^{-2}=k[0.10]^x[0.40]^y](https://tex.z-dn.net/?f=4.8%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5Ex%5B0.40%5D%5Ey)
(2)
Dividing 2 by 1 :![\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%7B1.2%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.20%5D%5Ey%7D)
therefore y=2.
b) From trial 2: (3)
From trial 3: ![9.6\times 10^{-2}=k[0.20]^x[0.40]^y](https://tex.z-dn.net/?f=9.6%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.20%5D%5Ex%5B0.40%5D%5Ey)
(4)
Dividing 4 by 3:![\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B9.6%5Ctimes%2010%5E%7B-2%7D%7D%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.20%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D)
, x=1
Thus rate law is ![Rate=k[A]^1[B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5E1%5BB%5D%5E2)
Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.
c) For calculating k:
Using trial 1: ![1.2\times 10^{-2}=k[0.10]^1[0.20]^2](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5E1%5B0.20%5D%5E2)
.
