Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
<span><span><span> </span><span>The strong forces oppose the electromagnetic force of repulsion between protons. Like ”glue” the strong force keeps the protons together to form the nucleus. </span>· The strong forces and electromagnetic forces both hold the atom together.</span><span> </span></span>
Answer:
All of the above.
Explanation:
In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.
When a solution is non ideal then it shows positive or negative deviation.
Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is 
and vapour pressure of component B is 
.
= Vapour pressure of component A in pure form
= Vapour pressure of component B in pure form
=Mole fraction of component A
=Mole fraction of component B
The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.
,
Therefore, 
Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.
Hence, option a,b,c and d are true.
<span>pb(no3)2 + 2 nacl → pbcl2 + 2nano3</span>
the most free particles are gases, then liquids, solids
