D. because the size and shape of the slope determines weather the load can be sustained and c is correct too because the shape affects how much you can load and b. is correct as also because if the volume is very large it can affect the rivers load and flow.
Answer: Selection proper
Explanation:
it's an anti-chance process, but subject to many constraints
thnxx for free point? ?????????
A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
the formula for calculating molality is:
molality = moles of solute/kilograms of solvent
1kg = 1000g so, 889g = 0.889kg
m = 1.43/0.889 = 1.61 molal
<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