Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
You need to know the energy frequency relationship for photons, which is thanks to Max Planck:
Photon Energy = Planck constant x Frequency
Rarranged:
Photon Energy / Planck Constant = Frequency
Planck Constant = 6.63x10^-34
2.93x10^-25 / 6.63x10^-34 = Frequency
Answer:
We need 4.28 grams of sodium formate
Explanation:
<u>Step 1:</u> Data given
MW of sodium formate = 68.01 g/mol
Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L
pH = 3.74
Ka = 0.00018
<u>Step 2:</u> Calculate [base)
3.74 = -log(0.00018) + log [base]/[acid]
0 = log [base]/[acid]
0 = log [base] / 0.42
10^0 = 1 = [base]/0.42 M
[base] = 0.42 M
<u>Step 3:</u> Calculate moles of sodium formate:
Moles sodium formate = molarity * volume
Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles
<u>Step 4:</u> Calculate mass of sodium formate:
Mass sodium formate = moles sodium formate * Molar mass sodium formate
Mass sodium formate = 0.063 mol * 68.01 g/mol
Mass sodium formate = 4.28 grams
We need 4.28 grams of sodium formate
3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
Hope this helps!!
Condensation
..when gas cool down into liquid statr its called condensation..