3.74×
3.74 × molecules of propane were in the erlenmeyer flask.
number of moles of propane can be calculated as moles of propane.
mass of propane = 0.274 g
molar mass of propane = 44.1
So this gives us the value of 6.21×
moles of propane
No one mole of propane As a 6.0-2 × 
so, 6.21 × × 6. 022 × 10^23
= 3.74 ×
Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 ×
<h3>What is erlenmeyer flask?</h3>
- A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
- It bears the name Emil Erlenmeyer after the German chemist.
<h3>What purpose does an Erlenmeyer flask serve?</h3>
- Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.
- For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.
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Answer:
The scientific method is a method of research in which a problem is identified, relevant data are gathered. Also hypothesis is made from this data, and the hypothesis is empirically tested.
Explanation:
Answer:
B) 2Crº + 6e- --> 2Cr3+
Explanation:
The process of oxidation is where electrons are lost. Thus, out of the 2 ions that change charge(Cr and Cu), we must choose the one where the oxidation number increases(which means electrons are lost). Cr goes from an oxidation number of 0 to an oxidation number of 3+, while Cu goes from an oxidation number of 2+ to 0. Thus, we are looking at the half reaction for Cr. Half reactions never have subtracting electrons, so the answer must be B. I am assuming that last plus should be a -->
Answer:
pH = 12.61
Explanation:
First of all, we determine, the milimoles of base:
0.120 M = mmoles / 300 mL
mmoles = 300 mL . 0120 M = 36 mmoles
Now, we determine the milimoles of acid:
0.200 M = mmoles / 100 mL
mmoles = 100 mL . 0.200M = 20 mmoles
This is the neutralization:
HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O
20 mmol 36 mmol 20 mmol
16 mmol
We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:
NaHCOO → Na⁺ + HCOO⁻
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹
These contribution of OH⁻ to the solution is insignificant because the Kb is very small
So: [OH⁻] = 16 mmol / 400 mL → 0.04 M
- log [OH⁻] = pOH → 1.39
pH = 14 - pOH → 12.61
