Answer:
The correct answer is 1.10.
Explanation:
Based on the given information, the molarity of the NaOH is 0.15 M, that is, 0.15 moles per liter of the solution.
Now the moles present in the 15 ml of the solution will be,
0.015 × 0.15 = 2.25 × 10⁻³ moles of NaOH or 0.0025 moles of NaOH
Now, molarity of the HNO₃ given is 0.20 M, which means 0.2 moles per liter of the solution.
Now the moles present in the 30 ml of the solution will be,
0.030 × 0.2 = 0.006 moles of HNO₃
Now the complete disintegration of acid and base will be,
NaOH (aq) (0.025 moles) ⇔ Na⁺ (aq) (0.025) + OH⁻ (aq) (0.025 moles)
HNO₃ (aq) (0.006 moles) ⇔ H⁺ (0.006 moles) + NO₃⁻ (aq) (0.006 moles)
Now the additional Hydrogen ions at titration point is,
= 0.006 - 0.0025 = 0.0035 moles of H+
Now the concentration of H+ ions in the 45 ml of the solution will be,
= 0.0035/45 × 1000
= 0.078 M
pH = -log[H⁺] = -log [0.078]
= 1.10
A. pOH=13
B. pOH=11.77
C. pOH=8.52
D. pOH=10.52
2. what is the pOH of a solution if the pH is 2.23
A. pOH=1
B. pOH=12
C. pOH=8
D. pOH=9
3. Given a solution of perchloric acid has a concentration of .023M, what is the pH and pOH
A. pH=2.3 and pOH=11.97
B. pH=1.64 and pOH= 12.36
C. pH=4.5 and pOH=9.95
The pH and pOH for a solution are related through the equation pH + pOH = 14. The pOH for solutions with a pH of 2.23 is 11.77, and in a solution of perchloric acid with a concentration of .023M, the pH is 1.64 and pOH is 12.36.
The concept of pOH and pH are related through the equation pH + pOH = 14 at 25 °C. If you have the value of one, you can easily find the other. For your first two questions, the pH of solution is given as 2.23. Therefore, the pOH would be 14 - 2.23 = 11.77. Your answer to both questions 1 and 2 is B. pOH=11.77.
For question 3, the perchloric acid has a concentration of .023M. To find the pH, we calculate -log[H3O+], which gives about 1.64. Therefore, the pOH would be 14 - 1.64 = 12.36. Your answer to question 3 is B. pH=1.64 and pOH=12.36.
#SPJ3
To find the concentration of X+ at a potential of 0.0610V, use the Nernst equation which describes the electrochemical potential of a system. Given the initial concentration of X+ and its potential, rearrange the equation to solve for the concentration of X+ at the new potential.
The problem given can be solved using the Nernst equation, which relates the reduction potential of an electrochemicalreaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and the activities of the chemical species undergoing the reduction.
The Nernst equation at 25 °C can be simplified as:
E = E° - (0.059/n) log [Cl^- /[X^+]
Where E is the electrode potential, E° is the standard electrode potential, n is the number of electron transferred and [Cl^- /[X^+] is the ratio of ion activities. Since the ion's activity coefficient is 1, we can treat [X^+] as the concentration of X^+.
If you apply this equation, using the given potentials and known initial concentration of X^+, you can solve for the concentration of X^+ when the potential is 0.0610V.
#SPJ12
The concentration of X+ can be calculated using the Nernst equation by substituting the initial and final potentials. By solving the equation, you can find the concentration of X+.
The concentration of X+ can be calculated using the Nernst equation. The Nernst equation relates the potential of a cell to the concentration of the ions involved.
The Nernst equation is given by:
E = Eº - (0.0592/n)log(Q)
Where E is the potential, Eº is the standard electrode potential, n is the number of electrons transferred, and Q is the reaction quotient.
In this case, the initial potential is 0.0460 V and the final potential is 0.0610 V. By substituting these values into the Nernst equation, you can solve for the concentration of X+.
#SPJ11
emperature. What is the independent variable in his experiment?
a. The unknown substance, because it's the only thing he changed
b.The temperature, because it's the only thing he changed
Answer:
a
Explanation:
Answer:
0,12 μmol/L of MgF₂
Explanation:
Preparation of solutions is a common work in chemist's life.
In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L
You have 0,00598 μmol but not Liters.
To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:
50,0 mL (1L/1000mL) = 0,05 L of water.
Thus, concentration in μmol/L is:
0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-
I hope it helps!
Answer:
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Explanation:
Step 1: Data given
ΔH = −320.1 kJ/mol
ΔS = −86.00 J/K · mol.
Step 2: Calculate the temperature
ΔG<0 = spontaneous
ΔG= ΔH - TΔS
ΔH - TΔS <0
-320100 - T*(-86) <0
-320100 +86T < 0
-320100 < -86T
320100/86 > T
3722.1 > T
The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)
We can prove this with Temperature T = 3730 K
-320100 -3730*(-86) <0
-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous
T = 3700 K
-320100 -3700*(-86) <0
-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous
The temperature is quite high because of the big difference between ΔH and ΔS.
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C