Answer:
Explanation:
0
All the charges of the protons are cancelled out by all the charges on the electrons.
Answer:
c. Atoms of two or more elements bond together
Explanation:
a chemical reactions forms products ( new substances); where it rearranges themselves to form new bonds.
KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which <span>then will react with 4-nitroaniline to form a diazonium salt.</span>
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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 Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [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>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Answer:
0.05
moles
Explanation:
In a mole of any substance, there exist
6.02⋅1023
units of that substance.
So here, we got:
3.01⋅1022Mg atoms⋅1mol6.02⋅1023M gatoms=0.05mol