Answer: The actions that must have affected the igneous rock in order to form the sedimentary rock is that (It must have been broken down into sediments).
Explanation:
Rocks are solid structures that occurs naturally which is made up of different minerals. There are three main types of rocks, these includes:
--> METAMORPHIC ROCKS: These are the type of rocks which are formed by temperature and pressure changes inside the Earth.
--> SEDIMENTARY ROCKS: these rocks are usually formed from pre-existing rocks through the process of weathering (breaking down) of rocks.
--> IGNEOUS ROCKS: these rocks are formed when molten magma cools beneath or above the earth surface.
The actions that must have affected the igneous rock in order to form the sedimentary rock is that the igneous rocks are broken down into smaller pieces by erosion and weathering processes. Sediments which are formed accumulates at the earth surface. Over a long period of time, these sediments builds successive layers on top of one another. The sediments near the base hardens to form sedimentary rocks. This justifies the statement as a correct option (It must have been broken down into sediments).
Answer:
35.4528731 amu
Explanation:
To appropriately get the atomic mass unit of chlorine, we can get the answer using the masses from the isotopes. This can be obtained as follows. What we do is that we multiply the percentage compositions by the masses.
Now let’s do this.
[75.77/100 * 34.969] + [24.23/100 * 36.966]
= 26.4960113 + 8.9568618 = 35.4528731
Answer:
30 protons
Explanation: Each zinc isotope contains 30 protons, 30 massive, positively charged nuclear particles
30 protons, 34 neutrons, and 30 electrons.
Answer:
An element that is oxidized is a reducing agent, because the element loses electrons, and an element that is reduced is an oxidizing agent, because the element gains electrons.
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.