1+14+(3*16)=63g/mol
hope it helps
Answer:
27.06 mL.
Explanation:
- Firstly, we need to calculate the molarity of acrylic acid.
Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.
M = (no. of moles of acrylic acid)/(V of the solution (L))
M = (mass/molar mass)acrylic acid / (V of the solution (L))
mass of acrylic acid = 0.0975 g, molar mass of acrylic acid = 72.06 g/mol, V of the solution = 250 mL = 0.25 L.
∴ M = (0.0975 g/72.06 g/mol)/(0.25 L) = 0.0054 M.
- For the acid-base neutralization, we have the role:
The no. of millimoles of acid is equal to that of the base at the neutralization.
<em>∴ (XMV) NaOH = (XMV) acrylic acid.</em>
X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),
M is the molarity.
V is the volume.
X = 1, M = 0.05 M, V = ??? mL.
X = 1, M = 0.0054 M, V = 250.0 mL.
∴ V of NaOH = (XMV) acrylic acid/(XM) NaOH = (1)(0.0054 M)(250.0 mL)/(1)(0.05 M) = 27.06 mL.
When a mixture of 10 moles of SO2 and 15 moles of O2 was passed over a catalyst, 10 moles of SO3 was formed.
PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-8.78) = 1.65*10^-9
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1.65*10^-9) = 6.06*10^-6
The concentration of OH- ions is 6.1*10^-6 M.
