<u>Answer:</u> The correct net ionic equation for the reaction is 
<u>Explanation:</u>
Net ionic equation is defined as the equations in which spectator ions are not included.
Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.
The balanced molecular equation is:

The complete ionic equation follows:

As calcium and sulfate ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:

Hence, the correct net ionic equation for the reaction is 
Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)

From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
The density of the metal is 3.457 g/cm³.
Explanation:
Given,
mass of metal = 121 grams
Volume = 35 cm³.
Density = ?
Density of the metal can be found by using the formula
Density = Mass/Volume
Substituting the values,
Density = 121/35 = 3.457 g/cm³.
Answer:
The equilibrium will go right, increasing the quantities of Y(g) and Z(s), when the amount of X(g) increases.
Explanation:
Equilibrium is a state in which the velocity of the formation of the products is equal to the velocity of the formation of the reactants. Thus, the concentrations remain constant.
By Le Chatêlier's principle, when the equilibrium is disturbed, the equilibrium must shift to reestablish the equilibrium. Thus, when X is put in the container, it intends to decompose and form Y and Z, thus, as higher is the initial X, as higher is the Y and Z formed.
1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?
Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.
2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.
3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.
What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.