Answer:
<h3>Step 1: </h3>
To relate the volume and molarity of a solution at two different concentrations, the expression used is :
M₁V₁ = M₂V₂
<h3>
step 2:</h3>
(2M) V₁ = (0.15M)(250ML)
V₁ = 18.75
step 3:
<h2>Pls, branliest! :)</h2>
<span>All of these. All the gases that are mentioned in each set identify with the law of multiple proportions since the all the compounds have oxygen ions in them. Law of multiple proportions is defined as formation of compound with oxide ions after the reaction of an element with oxygen.</span>
Answer:
Ka3 for the triprotic acid is 7.69*10^-11
Explanation:
Step 1: Data given
Ka1 = 0.0053
Ka2 = 1.5 * 10^-7
pH at the second equivalence point = 8.469
Step 2: Calculate Ka3
pKa = -log (Ka2) = 6.824
The pH at the second equivalence point (8.469) will be the average of pKa2 and pKa3. So,
8.469 = (6.824 + pKa3) / 2
pKa3 = 10.114
Ka3 = 10^-10.114 = 7.69*10^-11
Ka3 for the triprotic acid is 7.69*10^-11
Answer:
D. 108 grams of KNO3(Potassium nitrate)
Answer: 0.0826mol
PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
