Answer:

Explanation:
We are asked to find the density of a rock. Density is the substance's mass per unit volume. The formula for calculating density is:

The mass of the rock is 23 grams. The volume was found using water displacement. A known amount of water was measured, the rock was added to the graduated cylinder, and the water level was recorded again. The volume is the difference between the final and initial water level. The graduated cylinder originally had 20 milliliters and the water rose to 30 milliliters.
- volume = final water level - initial water level
- volume = 30 mL - 20 mL
- volume = 10 mL
Now we know the mass and the volume.
Substitute the values into the formula.

Divide.

The density of the rock is <u>2.3 grams per milliliter.</u>
Answer:
D. Reduction
Explanation:
In the electrolytic cell, reduction occurs at the cathode
The nobel prize in 1903 was given and awarded to Antoine Henri Becquerel, and Pierre Curie, Marie Curie, nee Sklodowska.
Reasons for the given awards;
Antoine Henri Becquerel because; "In recognition of the extraordinary services he has rendered by his discovery of spontaneous radioactivity."
And the others (Pierre Curie and Marie Curie, nee Sklodowska) because; "In recognition of the extraordinary services they have rendered by their joint researches on the radiation phenomena, discovered by Professor Henri Becquerel."
The answers is a, A series of steps used by scientists to investigate a hypothesis
Answer: The solubility product of AgCl is
.
Explanation:
The reaction equation is as follows.

Let us assume the concentration of
is 2S and concentration of
is S. Hence, the expression for
of this reaction is as follows.
![K_{sp} = [Ag^{+}]^{2}[CO^{2-}_{3}]\\8.2 \times 10^{-12} = (2S)^{2}(S)\\8.2 \times 10^{-2} = 4S^{3}\\S = 1.27 \times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCO%5E%7B2-%7D_%7B3%7D%5D%5C%5C8.2%20%5Ctimes%2010%5E%7B-12%7D%20%3D%20%282S%29%5E%7B2%7D%28S%29%5C%5C8.2%20%5Ctimes%2010%5E%7B-2%7D%20%3D%204S%5E%7B3%7D%5C%5CS%20%3D%201.27%20%5Ctimes%2010%5E%7B-4%7D)
This means that
is
. Now, the concentration of
is calculated as follows.
![[Cl^{-}] = \frac{mass}{molar mass}\\= \frac{0.003 g}{35.5 g/mol}\\= 8.45 \times 10^{-5} M](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B0.003%20g%7D%7B35.5%20g%2Fmol%7D%5C%5C%3D%208.45%20%5Ctimes%2010%5E%7B-5%7D%20M)
Hence,
for AgCl is calculated as follows.
Thus, we can conclude that solubility product of AgCl is
.