I guess the answer is MgO4S
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)
No it is an amino acid present in the synthesis of certain proteins. Elements are pure substances so an acid (which is a compound of different elements) would not be considered an element.
The density of the liquid is about 1.85 g/mL.
Density is mass/volume. The volume is given (45.2 mL). The mass must be found by subtracting the tare weight of the graduated cylinder from the total:
95.1 g- 11.4g = 83.7g
Using the mass and volume of the liquid, you can now calculate the density:
d = m/v = 83.7g/45.2 mL = 1.8517699115 g/mL.
Of the original values, the least number of significant figures are 3, so the answer must have a degree of accuracy of 3 significant figures:
1.8517699115 g/mL ≈ 1.85 g/mL.
