Dmitri Mendeleev
<span> Dmitri Mendeleev and Lothar Meyer individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass,
certain sets of properties recur periodically.</span>
when heat gained = heat lost
when AL is lost heat and water gain heat
∴ (M*C*ΔT)AL = (M*C*ΔT) water
when M(Al) is the mass of Al= 225g
C(Al) is the specific heat of Al = 0.9
ΔT(Al) = (125.5 - Tf)
and Mw is mass of water = 500g
Cw is the specific heat of water = 4.81
ΔT = (Tf - 22.5)
so by substitution:
∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)
∴Tf = 30.5 °C
Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
<h3>
Answer:</h3>
<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>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
