Answer:
about 10 pounds
Explanation:
your mass doesn't change but weight changes (due to gravity)
The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
Answer:
LiCl = 0.492 m
Explanation:
Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.
Our solute is lithium chloride, LiCl.
Our solvent is distilled water.
We do not have the mass of water, but we know the volume, so we should apply density to determine mass.
Density = mass / volume
Density . volume = mass
1 g/mL . 19.7 mL = 19.7 g
We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg
Let's determine the moles of LiCl
0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles
Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m
Answer: C) the values of Kb and Kw
Explanation: i just took the test
<h3>
Answer:</h3>
16.7 g 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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O