Answer: second option: 1.70 to 1.40
Explanation:
1) pH is defined using the formula pH = - log [H₃O⁺]
2) Given that the initial concentration is x and after doubling it becomes 2x, we calculate:
pHi = - logx
pHf = - log 2x = - log 2 - logx
Thus, pHf - pHi = - log2 - logx - (- logx) = - log2 ≈ - 0.30
⇒ pHi - pHf = 0.30, indicating that the final pH (with twice the hydronium ions) is 0.30 lower than the starting pH.
3) The only option that indicates a 0.30 decline in pH is the second one: from 1.70 to 1.40. Therefore, that is the correct choice.
Answer:

Explanation:
1. Molar concentration
Designate chloroform as C and acetone as A.
The molar concentration for C is derived from Moles of C per Litres of solution.
(a) Moles of C
We are assuming there are 0.187 moles of C.
This resolves that step.
(b) Litres of solution
Next, identify 0.813 moles of A.
(i) Mass of each component

(ii) Volume of each component

(iii) Volume of solution
Assuming mixing doesn't alter the total volume.
V = 15.08 mL + 59.70 mL = 74.78 mL
(c) Molar concentration of C

2. Molal concentration of C
Molal concentration is calculated as moles of solute per kilograms of solvent.
Total moles of C = 0.187 mol.
Mass of A = 47.22 g = 0.047 22 kg.

Answer:
The force is 38503.5N.
Explanation:
From the problem, we determine:
P (pressure) = 5.00 atm.
Next, to find the force in Newtons (N), we must convert 5 atm into N/m², as shown:
1 atm equals 101325 N/m².
So, 5 atm equals 5 x 101325 = 506625 N/m².
A (the piston area) = 0.0760 m².
Pressure signifies force per unit area, mathematically represented as
P = F/A.
From this, we find F = P × A.
F = 506625 × 0.0760.
Therefore, F = 38503.5N.
Thus, the piston experiences a force of 38503.5N.
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.
Answer:
Endothermic: water formation from ice and a cold instant ice pack Explanation: