Answer: If it was 3 mol of solute in 2 L of solution it would be 1.5 mol/L.
However when the solute dissolves in the water creating the solution, the volume increases. So 3 mol of solute in 2 L of water creates more than 2 L of solution.
The correct method for making a 3 mol/L solution would be to place some water into a two liter volume container. Dissolve all 3 mol of the solute into the water. Then add water to the 2 L mark. Now there is 3 mol of solute and 2 L of solution.
Explanation: I hope this helps XDDDD
Sulphur dioxide is greenhouse gas
may this will help you
Question:
a. a direct linear relationship
b. an inverse linear relationship
c. a direct nonlinear relationship
d. an inverse nonlinear relationship
Answer:
The correct option is;
d. An inverse nonlinear relationship
Explanation:
From the universal gas equation, we have;
P·V = n·R·T
Where we have the temperature, T and the number of moles, n constant, therefore, we have
P×V = Constant, because, R, the universal gas constant is also constant, hence;
P×V = C

Since P varies with V then the graphical relationship will be an inverse nonlinear as we have
V P C
1 5 5
2 2.5 5
3 1.67 5
4 1.25 5
5 1 5
6 0.83 5
7 0.7 5
8 0.63 5
9 0.56 5
10 0.5 5
Where:
V = Volume
P = Pressure
C = Constant = 5
P = C/V
The graph is attached.
The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.
<h3>KCl or NaF, which has a higher lattice energy?</h3>
The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.
<h3>MgO or CaO, which has a larger lattice energy?</h3>
MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.
To know more about compounds visit:
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The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
....(1)
Expression for rate law for second observation:
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)


Therefore, the initial rate of the reaction will be, 0.03 M/s