Answer:
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Explanation:
The equation for the reaction of trimethylamine when it is dissolved in water is:
C₃H₉N + H₂O ⇄ C₃H₉NH⁺ + OH⁻ Kb
1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.
It is a base, because it give OH⁻ to the medium
Expression for Kb (Molar concentration)
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Glycerol attractive forces are great than water. The harder to break, the more energy is needed.
Answer:
2 mol H₂O
Explanation:
With the reaction,
- 2H₂(g) + O₂(g) → 2 H₂O(g)
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
- 2.00 mol H₂ *
= 2 mol H₂O
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
Answer:
The flame will burn out
Explanation:
The oxygen will be kept out of the jar, without oxygen the flame cannot last long. The flame will eventually burn out. Nitrogen gas has no role to play in the burning. Nitrogen slows down the burning rate.