<h3>
Answer:</h3>
3.01 × 10²⁵ molecules H₂O
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
50.0 mol H₂O
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u />
= 3.011 × 10²⁵ molecules H₂O
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O
Answer: Where are the answers?
Explanation:
Explanation:
Each element in the periodic table has different but fixed number of the protons in nucleus of it's atom, which is known as the atomic number.
Transmutation of one chemical element into the another involves the changing of the atomic number. Such nuclear reaction requires millions of the times more energy as compared to normal chemical reactions. Thus, the dream of the alchemist of transmuting the lead into the gold was never achievable chemically .
Conversion of lead to gold in today's world:
This conversion is indeed possible. The requirements are a particle accelerator, tremendous supply of the energy. Nuclear scientists at the Lawrence Berkeley National Laboratory located in California, more than 30 years ago, succeeded in producing very minute amounts of the gold from the bismuth. Bismuth is a metallic element which is adjacent to the lead on periodic table. Same process would work for the lead but isolating gold at end of reaction would prove much more difficult because lead is available in many isotopes. The homogeneous nature of the element means that it is easier to separate the gold from the bismuth as compared to separate the gold from the lead which has four isotopic identities which all are stable.
It is topsoil, subsoil, and parent material
<u>Answer:</u> The value of equilibrium constant for the net reaction is 11.37
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u> ![A+2B\xrightarrow[]{K_1} 2C](https://tex.z-dn.net/?f=A%2B2B%5Cxrightarrow%5B%5D%7BK_1%7D%202C)
<u>Equation 2:</u> ![2C\xrightarrow[]{K_2} D](https://tex.z-dn.net/?f=2C%5Cxrightarrow%5B%5D%7BK_2%7D%20D)
The net equation follows:
![D\xrightarrow[]{K} A+2B](https://tex.z-dn.net/?f=D%5Cxrightarrow%5B%5D%7BK%7D%20A%2B2B)
As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:

We are given:


Putting values in above equation, we get:

Hence, the value of equilibrium constant for the net reaction is 11.37