Answer: -
C) because the moon rotates once on its axis in the same time that it takes to revolve around the earth
Explanation: -
The moon rotates about it's axis one full time at about the same time as it takes to rotate around the earth.
This leads to only one side of the moon always appearing to Earth, called the near side.
One side of the moon is always not seen from the earth. It is called the far side of the moon.
Thus we can never see the far side of the moon because the moon rotates once on its axis in the same time that it takes to revolve around the earth
Formation of a precipitate
A measure of the ability of a substance, or more generally of any physical system, to transfer heat energy to another physical system. The temperature of a substance is closely related to the average kinetic energy of its molecules.
I think-
<h3>
Answer:</h3>
2670 g Hg
<h3>General Formulas and Concepts:
</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:
</h3>
<u>Step 1: Define</u>
8.02 × 10²⁴ atoms Hg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Hg - 200.59 g/mol
<u>Step 3: Convert</u>
- Set up:

- Divide/Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2671.42 g Hg ≈ 2670 g Hg
Explanation:
It is given that the total volume is (10 mL + 60 mL) = 70 mL.
Also, it is known that
= 
Where,
= total volume
= initial volume
Therefore, new concentration of
= 
= 
= 0.43 M
New concentration of NaOH = 
= 
= 0.14 M
So, the given reaction will be as follows.

Initial: 0.43 0.14 0
Change: -0.14 -0.14 0.14
Equilibrium: 0.29 0 0.14
As it is known that value of
= 4.74
Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.
pH = ![pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 
= 4.74 + (-0.316)
= 4.42
Therefore, we can conclude that the pH of given reaction is 4.42.