Answer:
1)alkali metals
2)1g/ml
3) mercury
Explanation:
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<em><u>carryonlearing </u></em>
Answer:
C phosphorus has 5 bond sites.
Given:
The density of air = 1.19 g/L at 25°C and atmospheric pressure,
or
density = 1.19 x 10⁻³ kg/L
Volume of air in the room is
V = 12.5*19.5*6.0 = 1462.5 ft³
Note that
1 ft³ = 28.317 L
Therefore
V = (1462.5 ft³)*(28.317 L/ft³) = 4.1414 x 10 ⁴ L
By definition, mass = density*volume.
Therefore, the mass is
(1.19 x 10⁻³ kg/L)*(4.1414 x 10⁴ L) = 49.283 kg
Answer: 49.3 kg (nearest tenth)
Answer: 78.125%
Explanation:
From the question, we are informed that a student obtained 2.00g of pure caffeine following recrystallization from a crude sample, which originally weighed 2.56g.
The percent recovery of pure caffeine from crude will be calculated as:
= Mass of pure caffeine / Mass of crude sample × 100
= 2.00 / 2.56 × 100
= 78.125%
<h3>
Answer:</h3>
2900 g Au
<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 Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 14.7 mol Au
[Solve] g Au
<u>Step 2: Identify Conversion</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2895.46 g Au ≈ 2900 g Au