Yes it can. i hope i helped
Answer:
first of all is that really chemistry
Answer:
- Volume = <u>2.0 liter</u> of 1.5 M solution of KOH
Explanation:
<u>1) Data:</u>
a) Solution: KOH
b) M = 1.5 M
c) n = 3.0 mol
d) V = ?
<u>2) Formula:</u>
Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:
<u>3) Calculations:</u>
- Solve for n: M = n / V ⇒ V = n / M
- Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter
You must use 2 significant figures in your answer: <u>2.0 liter.</u>
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NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
