The answer is b because I learned this in 5th grade and I got it right
<h3>
Answer:</h3>
5.89 × 10^23 molecules of F₂
<h3>
Explanation:</h3>
The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;
5F₂ + 2NH₃ → N₂F₄ + 6 HF
We are given 66.6 g NH₃
We are required to determine the number of fluorine molecules
<h3>Step 1: Moles of Ammonia </h3>
Moles = Mass ÷ Molar mass
Molar mass of ammonia = 17.031 g/mol
Moles of NH₃ = 66.6 g ÷ 17.031 g/mol
= 3.911 moles
<h3>Step 2: Moles of Fluorine </h3>
From the equation 5 moles of Fluorine reacts with 2 moles of ammonia
Therefore,
Moles of fluorine = Moles of Ammonia × 5/2
= 3.911 moles × 5/2
= 9.778 moles
<h3>Step 3: Number of molecules of fluorine </h3>
We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules
Therefore;
1 mole of F₂ = 6.022 × 10^23 molecules
Thus,
9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole
= 5.89 × 10^23 molecules
Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules
Answer:
Newton
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Answer:
maybe 1 or 3 im not sure
Explanation:
i didn't study it yet sorry for not helping but try asking someone else
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .230 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.230-x ≈ 0.230
4.5x10^-4 = x^2/0.230
Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.01M/0.230M = .0434 or
≈4.34% dissociation.
