Answer:
Explanation:
1) Molarity of the solution is (0.50/0.75)x1 = 0.66666 Mol/L
2) Molar mass of NaCl is 23+35 = 58g
Molarity = (0.5/58)/0.075 = 0.115 Mol/L
3) Molar mass of
= 7x2 + 32 + 16x2 = 78g
Molarity = (734/78)/0.875 = 10.754 Mol/L
5) Molar mass of

= 207.2 + ((12x2) + (1x3) + (16x2))x4
= 207.2 + (24+3+32)x4 = 443.2g
Molarity = (0.0672/443.2)/0.1335 = 0.001135 Mol/L
6) Both are the same.
Solution 1 Molarity = 1.0/1.0 = 1 Mol/L
Solution 2 Molarity = 1.0/1.0 = 1 Mol/L
Molarities of the solutions are the same.
Radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration or nuclear disintegration) is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay (-decay), beta decay (-decay), and gamma decay (-decay), all of which involve emitting one or more particles or photons. The weak force is the mechanism that is responsible for beta decay, while the other two are governed by the usual electromagnetic and strong forces.[1]
Answer:- Molarity of the acid solution is 0.045M.
Solution:- The balanced equation for the reaction of given acid and base is:

From the balanced equation, they react in 1:1 mol ratio. So, we could easily solve the problem using the equation:

where,
is the molarity of acid,
is the molarity of base,
is the volume of acid and
is the volume of base.
Let's plug in the given values in the equation:

on rearranging the above equation:

= 0.045M
So, the molarity of the acid solution is 0.045M.
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.