I can tell you one. Must have particles
Answer:
7.07
Explanation:
HA = weak acid = 0.053
A+ = conjugate base = 0.045
Ka = 7.2x10^-8
Ka = [H+][A-]/HA
7 2x10^-8 = [H+][0.045]/0.053
[H+] = 7.2x10^-8 x 0.053/0.045
= 8.48x10^-8
PH = -log[H+]
= -log[8.48x10^-8]
PH = -[login.48 + log10^-8]
PH = -0.928 - (-8)log10
= 7.07
Atomic mass Ca = 40.078 a.m.u
40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??
72.8 x ( 6.02x10²³) / 40.078 =
4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms
hope this helps!
No, actually adawadawada and awawawaw usually addawadadaw but also awawawa so it’s a possibility but very rare.
Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
