Answer:
n₂ = 2.55 mol
Explanation:
Given data:
Initial number of moles = 0.758 mol
Initial volume = 80.6 L
Final volume = 270.9 L
Final number of moles = ?
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = initial number of moles
V₂ = Final volume
n₂ = Final number of moles
now we will put the values in formula.
V₁/n₁ = V₂/n₂
80.6 L / 0.758 mol = 270.9 L/ n₂
n₂ = 270.9 L× 0.758 mol / 80.6 L
n₂ = 205.34 L.mol /80.6 L
n₂ = 2.55 mol
In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.
Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.
Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M
The concentration of HBr is 0.127 M.
Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,

where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:


The relation between the equilibrium constant and standard Gibbs free energy is:

where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:


Therefore, the value of equilibrium constant for this reaction at 262.0 K is 
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
I need to find the x pls help and show work 73838