<h2>
Answer:</h2>
390 g KNO₃
<h2>
General Formulas and Concepts:</h2><h3><u>Chemistry</u></h3>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3><u>Math</u></h3>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h2>
Explanation:</h2>
<u>Step 1: Define</u>
2.3 × 10²⁴ formula units KNO₃
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g.mol
Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
<u>Step 3: Convert</u>
<u />
= 386.172 g KNO₃
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules and round.</em>
386.172 g KNO₃ ≈ 390 g KNO₃
Answer:
Explanation:
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.
Hello!
The mass number in isotope notation is denoted A, the atomic number is denoted as Z, and the element is denoted as X.
In the given isotope, the mass of the isotope is 212 amu, and the atomic number is 82.
We know that the number of electrons, and protons are equal to the atomic number. Therefore, there are 82 protons. Also, to find the number of neutrons, we subtract the atomic number from the atomic mass.
212 - 82 = 130 neutrons
<u>Final answers</u>:
- Atomic Number: 82
- Mass number: 212
- Number of Protons: 82
- Number of Neutrons: 130
Answer:
Chloroform= limiting reactant
0.209mol of CCl4 is formed
And 32.186g of CCl4 is formed
Explanation:
The equation of reaction
CHCl3 + Cl2= CCl4 + HCl
From the equation 1 mol of
CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4
From the question
25g of CHCl3 really with Cl2
Molar mass of CHCl3= 119.5
Molar mass of Cl2 = 71
Hence moles of CHCl3= 25/119.5 = 0.209mol
Moles of Cl2 = 25/71 = 0.352mol
Hence CHCl3 is the limiting reactant
Since 1 mole of CHCl3 gave 1mol of CCl4
It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4
Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g
