Answer:
friend according to the question the answer is +3 ....
Methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Answer:
Let the mixture is X% by mass of CuSO
4
.5H
2
O and 100 - X % by mass of MgSO
4
.7H
2
O. 5.0 g of mixture will contain 0.05X g CuSO
4
.5H
2
O and 5.0 - 0.05X g MgSO
4
.7H
2
O
The molar masses of CuSO
4
.5H
2
O and MgSO
4
.7H
2
O are 249.7 g/mol and 246.5 g/mol respectively.
The number of moles of CuSO
4
.5H
2
O=
249.7
0.05X
=2.00×10
−4
X moles.
Explanation:
Pls mark it as branliest answere thanks
Answer: Rate law=
, 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)
From trial 2:
(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:
(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)
.