First, we have to get moles of CH3COONa = mass/molar mass
= 20 g / 82.03 g/mol = 0.244 moles
when we have [CH3COOH] = 0.15 M
∴ [CH3COONa] = moles of CH3COONa / Volume of solution
= 0.244 moles / 0.5L = 0.488 M
when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5
So we can get Pka = -㏒Ka
= -㏒(1.8 x10^-5)
= 4.7
now we will use Henderson - Hasselbalchn equation to get the PH:
PH = Pka + ㏒[conjugate basic/weak acid]
when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:
PH = 4.7 + ㏒ (0.488/0.15)
= 5.2
b) when we have this equation for the reaction:
HCl + CH3COONa → CH3COOH + NaCl
ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-
when HCl + H2O → H3O+ + Cl-
∴ the reaction will be:
CH3COO- (aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
Answer:
a liquid
Explanation:
I'm not for sure but I just looked it up
Given:
The density of air = 1.19 g/L at 25°C and atmospheric pressure,
or
density = 1.19 x 10⁻³ kg/L
Volume of air in the room is
V = 12.5*19.5*6.0 = 1462.5 ft³
Note that
1 ft³ = 28.317 L
Therefore
V = (1462.5 ft³)*(28.317 L/ft³) = 4.1414 x 10 ⁴ L
By definition, mass = density*volume.
Therefore, the mass is
(1.19 x 10⁻³ kg/L)*(4.1414 x 10⁴ L) = 49.283 kg
Answer: 49.3 kg (nearest tenth)