Answer : The rate for a reaction will be 
Explanation :
The balanced equations will be:

In this reaction,
and
are the reactants.
The rate law expression for the reaction is:
![\text{Rate}=k[A]^2[B]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D%5E1)
or,
![\text{Rate}=k[A]^2[B]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D)
Now, calculating the value of 'k' by using any expression.
![\text{Rate}=k[A]^2[B]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D)


Now we have to calculate the initial rate for a reaction that starts with 1.48 M of reagent A and 1.32 M of reagents B.
![\text{Rate}=k[A]^2[B]^0[C]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D%5E0%5BC%5D%5E1)


Therefore, the rate for a reaction will be 
Answer:
When you place an iron nail in a basin of water and leave it for two days, what will happen to the iron nail? It will form rust on the surface of the nail. Rust forms when iron combines with oxygen to form iron oxide.
Explanation:
Assuming they were clean of rust to begin with, and are made of an ordinary steel - they start to rust - will turn orange, and the orange sediment...
Not much if any while it's in the water for that period of time. However once you remove it then it will be exposed two oxygen and will start to ox...
If it is not coated properly It will form Rust .
Answer:
a) 129.14 g/mol
b) 8.87
Explanation:
Given that:
mass of the ionic compound [NaA] = 18.08 g
Volume of water = 116.0 mL = 0.116 L
Let the mole of the acid HCl = 0.140 M
Volume of the acid = 500.0 mL = 0.500 L
pH = 4.63
= 1.00 L
Equation for the reaction can be represented as:

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L
= 0.140 mol
Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium
Thus, the molar mass of the sample = 
= 129.14 g/mol
b) since pH = pKa
Then pKa of HA = 4.63
Ka = ![10^{-4.63]](https://tex.z-dn.net/?f=10%5E%7B-4.63%5D)
= 
![[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}](https://tex.z-dn.net/?f=%5BA%5E-%5Dequ%20%3D%20%5Cfrac%7B0.140M%2A1.00L%7D%7B1.00L%2B0.116L%7D)

= 0.1255 M
of HA = 


+
+ 
Initial 0.1255 0 0
Change - x + x + x
Equilibrium 0.1255 - x x x
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
![4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}](https://tex.z-dn.net/?f=4.35%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B%5B0.1255-x%5D%7D)
As
is very small, (o.1255 - x) = 0.1255

[OH⁻] = 
But pOH = - log [OH⁻]
= - log [
]
= 5.13
pH = 14.00 = 5.13
pH = 8.87