Molarity is the estimation of the solute concentration in the solution. The final molar concentration of the acetate anion in the solution is 0.006 M.
<h3>What is Molarity?</h3>
Molarity is the ratio of the moles to the volume in liters.
Given,
Mass of lead (II) acetate = 0.258 gm
Volume of solution = 0.25 L
Moles are calculated as:
Moles = mass ÷ molar mass
= 0.258 g ÷ 325.29 g/mol
= 0.00079
Molarity is calculated as:
M = moles ÷ volume
= 0.00079 ÷ 0.25
= 0.0032
The molar concentration of the acetate ions: 2 × 0.0032 = 0.0064 M
Therefore, the molarity of the acetate anion in the solution is 0.006 M.
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Answer:
Theoretical 32.04 g
The precent yeild 87. 39 %
Explanation:
Mole of H2 = m/mw = 4/2 = 2 mol
From the equation
1 mol H2 --> 1/2 mol CH3OH
2 mol H2 -->??
So we have 1 mol Ch3oh
Mass of Ch3oh = mole *Mw = 1* 32.04
= 32.04g >>> the theoretically
For percent yeild = actual / theoretical * 100%
( 28.0/32.04) *100% =
87.93%
Answer:
Do your legs hurt from running ♂️ through my mind all night?
Explanation:
Did you just come out of the oven? Because you're hot
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
