This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
Delta T= T final - T initial
Tfinal= -101.1 °C
Tinitial= -0.5 °C
•Delta T = -101.1°C - (-0.5°C)
=100.6°C
Kelvin= °C + 273
= -100.6 + 273
= 172.4 Kelvin
Explanation:
nnbbnmkmknn bnnnbbtbbbbn' nn' t
Answer:
Zn + 2HCl → ZnCl2 + H2
Explanation:
Zn + HCl → ZnCl2 +
The complete equation is given below:
Zn+ HCl → ZnCl2 + H2
Now we can balance the equation by doing the following:
There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:
Zn + 2HCl → ZnCl2 + H2
The molar mass of B(NO₃)₃ - Boron nitrate : 196.822 g/mol
<h3>Further explanation</h3>
In stochiometry therein includes
<em>Relative atomic mass (Ar) and relative molecular mass / molar mass (M) </em>
So the molar mass of a compound is given by the sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
The molar mass of B(NO₃)₃ - Boron nitrate :
M B(NO₃)₃ = Ar B + 3. Ar N + 9.Ar O
M B(NO₃)₃ = 10.811 + 3. 14,0067 + 9. 15,999
M B(NO₃)₃ = 196.822 g/mol
