Explanation:
Mole = 0.5 = n/NA
n = no. of molecule
NA = avogadro number = 6.023x 10^23
so
N/NA = 0.5
N = 0.5 x 6.023 x 10^23
N = 3.0115 x 10^23
Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
Balanced reaction-:
<h3>2AgNO3(aq)+K2CrO4(aq)→Ag2CrO4(s)+2KNO3(aq)</h3>
Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol
=1.589⋅10^−7 mol
Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M
Ksp of Ag2CrO4
=[Ag+]2[CrO42−]
1.2⋅10−12=[2s]2[s]
4s3=1.2⋅10−12
s=6.69⋅10−5 M
Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
<h3>What is the molarity calculation formula?</h3>
The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.
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2300 dL = 230 L
<span>23 L = 23 L </span>
<span>230 cL = 2.3 L </span>
<span>230000 mL = 230 L </span>
<span>2300000 nL = 0.0023 L </span>
<u>Answer:</u> The rate law expression is
and value of 'k' is 
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:

Rate law expression for the reaction:
![\text{Rate}=k[NO]^a[O_2]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5Ea%5BO_2%5D%5Eb)
where,
a = order with respect to nitrogen monoxide
b = order with respect to oxygen
- <u>Expression for rate law for first observation:</u>
....(1)
- <u>Expression for rate law for second observation:</u>
....(2)
- <u>Expression for rate law for third observation:</u>
....(3)
Dividing 1 from 2, we get:

Dividing 1 from 3, we get:

Thus, the rate law becomes:
![\text{Rate}=k[NO]^2[O_2]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO%5D%5E2%5BO_2%5D%5E1)
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
![8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}](https://tex.z-dn.net/?f=8.55%5Ctimes%2010%5E%7B-3%7D%3Dk%5B0.030%5D%5E2%5B0.0055%5D%5E1%5C%5C%5C%5Ck%3D1.727%5Ctimes%2010%5E3M%5E%7B-2%7Ds%5E%7B-1%7D)
Hence, the rate law expression is
and value of 'k' is 