Answer:
(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)
Explanation:
A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.
Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.
The balanced equation will read:
Pb(NO3)2 (aq) + LiSO4 (aq) ➡️ PbSO4 (s) + 2️⃣LiNO3 (aq)
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Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
We calculate first for the number of moles of gases in the sample through the ideal gas equation.
n = PV/RT
n = (725 mmHg/760 mmHg/atm)(0.255 L) / (0.0821 L.atm/mol.K)(65 + 273.15)
n = 8.76 x 10^-3 mol
Then, we calculate for the mol N2 using the ratio of the pressure.
n N2 = (8.76 x 10^-3 mols)(231 mmHg/725 mmHg)
n N2 = 2.79 x 10^-3 moles
Then, multiply the value with the molar mass of N2 which is 28 grams per mol giving us the answer of 0.078 grams.