Answer:
The rate law for the decomposition reaction is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be 
Explanation:

The rate law can be expressed as:
..[1]
When the drug concentration is tripled, the decomposition rate rises by a factor of nine.
![[D]'=3[D]](https://tex.z-dn.net/?f=%5BD%5D%27%3D3%5BD%5D)

...[2]
[1] ÷ [2]
![\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7BR%27%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5BD%27%5D%5Ex%7D)
![\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B9R%7D%3D%5Cfrac%7Bk%5BD%5D%5Ex%7D%7Bk%5B3D%5D%5Ex%7D)

Solving for x results in:
x = 2.
This indicates a second-order reaction.
The decomposition reaction's rate law is:
![R=k[D]^2](https://tex.z-dn.net/?f=R%3Dk%5BD%5D%5E2)
The unit for the rate constant will be:
![k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BR%7D%7B%5BD%5D%5E2%7D%3D%5Cfrac%7BM%2Fs%7D%7B%28M%29%5E2%7D%3DM%5E%7B-1%7Ds%5E%7B-1%7D)
The unit for the rate constant will be
.
Sagot:
0.1 M NaCl
Paliwanag:
Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.
Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;
Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)
Ang puting precipitate na nabuo ay PbCl2.
Response:
Here's my calculation
Clarification:
Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We need to determine the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's compile all the information in one location.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial values
The graph below visualizes the initial concentrations as plotted on the vertical axis.
Solution:
Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.
Thus,
Molality = moles of solute / kg of solvent.
Therefore, kg of solvent = moles of solute / molality.
moles of solute = mass / molar mass
= 25.31 g / 101.1 g/mole
= 0.2503 mole.
kg of solvent = 0.2503 mole / 0.1982 m
= 1.263 kg
= 1263 g.
This is the final answer.