<h3>
Answer:</h3>
128 g HCl
<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>
<u>Stoichiometry</u>
- Reaction Mole Ratios
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)
↓
[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)
[Given] 3.25 mol Mg
[Solve] x g HCl
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol HCl
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol
<u>Step 3: Stoich</u>
- [S - DA] Set up:
- [S - DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
127.61 g HCl ≈ 128 g HCl
First, you need to know 1 kg = 10^3 g. And 1 m^3 = 10^6 m^3. So the 1 g/cm3 = 10^3 kg/m3. So the answer is 1.93*10^4 kg/m3.
Answer:
Cada uno de esos orbitals sons los differentes grupos en la tabla periodica.
Explanation:
Answer:
There are 5.43 grams of NaOCl
Explanation:
The given percent by mass of the solute tells us that out of the 150 g of the solution, 3.62% are due solely to the solute.
In other words, <u>the mass of the solute in the solution is</u>:
- 150 g * 3.62/100 = 5.43 g
Thus, in 150 grams of the given bleach solution, there are 5.43 grams of sodium hypochlorite.