Answer:
nilon is inorganic so they can be in the soil up to 5000 years
Explanation:
pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.
pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.
pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.
Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.
Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :
HCL(0.040M)----> H+(0.040M) +CL-(0.040M)
HBr(0.075M)----> H+(0.075M) +Br-(0.075M)
so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.
pH is calculated as:
pH = -log[H+]
Substituting values in the equation:
log(0.115M)= 0.939 pH
pOH is calculated as:
14 - pH = pOH
Substituting values in the equation above:
14 - 0.939= 13.061 pOH
Therefore, pH is 0.939 and pOH is 13.061.
Learn more about pH and pOH here:
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Correct Answer: <span>(2) The mixture is homogeneous and cannot be separated by filtration.
Reason: Solution is said to be homogenous, if it contains single phase. In present case, salt and water, when dissolved in water forms a single phase. Hence, it is referred as homogenous solution. Also, individual components of homogenous system cannot be separated by the process of filtration. </span><span />
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
