Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
We can use the ideal gas equation:
PV = nRT
P = 202.6kPa = 202600 Pa (You have to
multiply by 1000)
n = 0.050 mole
R = 0.082 atm*l/(K*mol)
T = 400K
We will have to convert from Pa to atm or
viceversa.
101325 Pa________1 atm
202600 Pa________x = 2.00 atm
2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K
V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm
= 0.82 liters = 820 mililiters
Answer:
3. 75.0%
Explanation:
2 ClO2(g) + F2(g) → 2 FClO2(g)
First order with respect to ClO2 and F2.
This means the rate equation is given as;
Rate = k [ClO2][F2]
When the initial concentrations of ClO2 and F2 are equal?
Let's assume an initial value of 1 for both reactants, so rate equation is given as;
Rate = k * 1 * 1 = k
The rate after 25% of the F2 has reacted is what percent of the initial rate?
The concentration left of F2 is 75% ( 100% - 25%) = 0.75
Concentration of ClO2 remains 1.
So rate equation is given as;
Rate = k * 1 * 0.75 = 0.75 k
Comparing 0.75k and k.
This means our answer is;
3. 75.0%
Explanation:
equations to note:
density= mass/volume
mass= volume *density
volume= mass/density
you have a volume- 8.33cm3
you have a density- 2.07 g/cm3
Answer:
8.33cm3 * 2.07g/cm3= 17.24g
mass= 17.24g
The balanced chemical reaction would be:
KHC8H4O4<span> (aq) + </span>NaOH<span> (aq) → NaKC8H4O4 (aq) + H2O.
The concentration of the NaOH is equal 0.1 M. We use this and the volume given above to determine the mass of KH</span>C8H4O4. We do as follows:
0.1 mol / L NaOH (.015 L) ( 1 mol KHC8H4O4 / 1 mol NaOH) (204 g / 1 mol) = 0.306 g KHC8H4O4
