Answer:
24.9%
Explanation:
According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:
0.0927 × 100 = 9.27%
Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;
100% - 9.27% = 90.73%
THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles
N.B: mole = mass/molar mass
Given the Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
For NaCl;
0.0927 = mass/58.44
mass = 0.0927 × 58.44
5.42g
For H2O;
9.073 = mass/18.016
mass = 9.073 × 18.016
= 16.35g
Total mass of solution = 16.35g + 5.42g = 21.77g
Mass percent of NaCl = mass of NaCl/total mass × 100
% mass of NaCl = 5.42g/21.77g × 100
= 0.249 × 100
= 24.9%
Answer:
The reaction rate becomes quadruple.
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
Thus,
Given that:- The rate law is:-
![r=k[A_2][B_2]](https://tex.z-dn.net/?f=r%3Dk%5BA_2%5D%5BB_2%5D)
Now, ![[A'_2]=2[A_2]](https://tex.z-dn.net/?f=%5BA%27_2%5D%3D2%5BA_2%5D)
and ![[B'_2]=2[B_2]](https://tex.z-dn.net/?f=%5BB%27_2%5D%3D2%5BB_2%5D)
So, ![r'=k[A'_2][B'_2]=k\times 2[A_2]\times 2[B_2]=4\times k[A_2][B_2]=4r](https://tex.z-dn.net/?f=r%27%3Dk%5BA%27_2%5D%5BB%27_2%5D%3Dk%5Ctimes%202%5BA_2%5D%5Ctimes%202%5BB_2%5D%3D4%5Ctimes%20k%5BA_2%5D%5BB_2%5D%3D4r)
<u>The reaction rate becomes quadruple.</u>
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
