Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:
Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
Answer:
510 g NO₂
General Formulas and Concepts:
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
- Reading the Periodic Table
- Writing Compounds
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u />
= 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂
I have a feeling that #2 is the one.
These answers dont make sense
1. the 2s2 orbital will give one of its electrons to the 2p5 orbital so the configuration would be 1s22s12p6 (2s1 is half filled and 2p6 is completely filled which is a much more stable configuration)
2. Neon does not need to ionize it is a noble gas
