Answer: 0.4mL of the stock solution is needed
Explanation:
This is simply a dilution analysis. From the question, the following were obtained:
C1 = 25 mg/mL = 25x10^-3g/mL
V1 =?
C2 = 25 ug/mL = 25x10^-6g/mL
V2 = 400ml
Applying the dilution formula, we have:
C1V1 = C2V2
25x10^-3 x V1 = 25x10^-6 x 400
Divide both side by 25x10^-3, we have:
V1 = (25x10^-6 x 400) / 25x10^-3
V1 = 0.4mL
Therefore, 0.4mL of the stock solution need.
Answer:
For more information about each chemical, view the safety data sheets. Click the Product Resources tab followed by Datasheet. You’ll need these materials: test tube rack test tubes (quantity: 6) test tube labels graduated cylinder wash bottle with distilled water (Tap water is acceptable, but it can skew the experimental results.) pipettes (quantity: 2) forceps 50-milliliter Erlenmeyer ask scoop test tube brush apron goggles gloves pen or re-tip marker small piece of sandpaper chemicals from Edmentum Lab kit: copper metal strip (quantity: 1) iron nails (quantity: 2) zinc metal strips (quantity: 3) copper(II) sulfate solution (10 milliliters) iron(III) nitrate solution (10 milliliters), also called ferric nitrate magnesium sulfate solution (10 milliliters) hydrochloric acid (30 milliliters) sodium bicarbonate (7.5 grams
Hope this helps
Explanation:
Answer:
17 g Ba(NO₂)₂
General Formulas and Concepts:
<u>Chemistry</u>
- Stoichiometry
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
4.5 × 10²² molecules Ba(NO₂)₂
<u>Step 2: Define conversion</u>
Molar Mass of Ba - 137.33 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Ba(NO₂)₂ - 137.33 + 2(14.01) + 4(16.00) = 229.35 g/mol
<u>Step 3: Dimensional Analysis</u>
<u />
= 17.1384 g Ba(NO₂)₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
17.1384 g Ba(NO₂)₂ ≈ 17 g Ba(NO₂)₂
Oxygen can combine with a metal to produce a compound
Answer:
Zirconium has 40 protons and 51 neutrons
