[Chemistry] Determine the amount of fluorine?
Good afternoon,
Unfortunately, I don't really understand the following task yet.
I'm really looking forward to your explanations so that I can finally understand it perfectly.
Task 3
a) What amount of fluorine (in g) occupies a volume of 6 liters under standard conditions?
b) Explain the reaction of magnesium with oxygen according to Dalton's atomic theory.
c) Give the enthalpy diagram (with technical terms) for b).
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Good Morning,
(a)
under normal conditions, one mole of a gas enters 22.4 L. This allows you to determine the molar amount of 6 L of fluorine gas:
1 mol = 22.4 L
x moles = 6.0 L
1 mol to 22.4 L = x mol to 6.0 L
1 mol • 6.0 L ÷ 22.4 L = x mol
0.27 mol = x mol
And then you can talk about the relationship
m [g] = n [mol] • M [g/mol]
calculate the mass of the fluorine:
m [g] = 0.27 mol • 38 g/mol
m [g] = 10.26 g
Answer:
The mass of fluorine gas, which takes up a volume of 6 liters under normal conditions, is about 10.26 g.
(b)
Magnesium is a metal composed of individual magnesium atoms. Oxygen is a smallest non-metal, whose particles are two-atomic minimolecules, but at the time when Dalton had set up his atomic hypothesis, they did not yet know. Therefore, Dalton (wrongly) would have assumed that it would consist of individual atoms also in the case of oxygen.
And then Dalton would have described this: A substance portion of x atoms magnesium and x atoms oxygen connects to x atomic groups magnesium oxide.
For example, he would have shown that the "law of the preservation of the masses" (Lavoisier, 1789) and the "law of constant proportions" (Proust, 1797).
16 g of oxygen with 24 g of magnesium would react to 40 g of magnesium oxide (ie the masses of the two reactants are always in a ratio of 60% Mg: 40% O).
What kind of bonds are (ion binding) as they arise (ion formation by electron transfer) or why the two elements even connect to one another, the good Dalton had no idea.
(c)
The energy diagram of an exothermic reaction…
The flame of a burning piece of magnesium releases enormous amount of energy (strong exothermic reaction). The flame is hot and glowing between 2,400 and 3,000 degrees Celsius (white glow; strong release of light).
LG from the Waterkant
Thank you for your detailed answer, I might have a few questions later if you (here tomorrow) want to answer them. Greetings to the North Sea Coast(?), a happy new year and good luck, success and health! 🍀
Is that so with each gas that under normal conditions one mole of a gas is 22.4 L? If so, why is this the case? What should you know about “Mol”?
Yes and no… The 22.414 L applies to a so-called "ideal gas." For real gases, the exact value actually depends on the substance. But even for real gases, it's always around 22.4 L under normal conditions. That's why I used this value for my calculations.
This is because, at a given pressure, temperature, and number of particles, the gas occupies the same volume. In other words: equal volumes (at constant pressure and temperature) can accommodate the same number of particles of a gas.
If you are interested in more details:
https://www.chemie.de/lexikon/Molares_Volumen.html
What you should know about the mole is that it's the unit of a quantity of substance. It ultimately indicates how many particles are contained in the portion of substance. The "particles" can be anything, such as cars, people, eggs, etc., but in chemistry, it's usually atoms, ions, molecules, electrons, etc.
It is true that 1 mole of anything always contains 6.022 • 10 23 particles. It is therefore a fixed number, much like a dozen always means 12 particles.
Important connections (around the mole) are
Number of particles N : N = N A • n
or
Mass m (of an amount of substance n ): m = n • M
or
Molar volume: V m = V ÷ n
Greetings from the waterfront
I still have a few questions about your answer to task b) (explain the sequence of the reaction of magnesium with oxygen to Dalton's atom theory).
Questions:
Magnesium would have interpreted Dalton correctly, right? Only in the case of oxygen would he have said that it consists of an atom and not, as we know today, of two-atom minimolecules, right?
Is that still true for today's knowledge?
Does this also belong to the answer, since it was only asked to explain the course of the reaction to Dalton's atomic theory? Therefore, the >>Act of the Conservation of Mass < < < must also be mentioned here?
I have never heard anything about bindings such as >> ion binding < < < < or >> ion formation by electron transfer < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < < > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > > > > > > >
Yes and yes.
Nope. We would express it like this today:
A substance portion of 2x atoms magnesium and x oxygen minimolecules combine to form 2x magnesium oxide formulae units.
Not necessarily, but I think Dalton would have mentioned it in connection with his nuclear hypothesis (to support his explanation)…
If you haven't had the ion bond yet, you don't have to know it. Like Dalton. He didn't know her during his lifetime.
Good evening, DedeM,
Thanks again for your perfect help with chemistry! 🤩
If you can do math so well and have the pleasure and time to help me with a small task, then you would like to have the following text:
I urgently need help with a math task at probabilities. I posted in addition to the question (other question, see link below) the calculation path with the result of my teacher and above in the question is my calculation. The answerers and also I come to a different outcome than my teacher. I'm afraid I don't understand my teacher's path. It would help me a lot if you looked at the question and told me how my teacher calculated it and whether my bill or my teacher's bill is correct.
Here you get to the question:
Love
2002
Good lunch, if you like later, you can look at my latest question if you like. It's about recruiting Mol. So with the formula M=m/n. The calculations have so far been used only for elements but not for connections. There I asked the following question: “How many moles are 63 g of water?”. Here you could ask:
Love
Thank you very much! I now create a card so I can remember it well 🙏
Yeah, basically. But it does not always have to react elements with each other. It can also be connections, which is why your introduction with "Normally stand left … yes always the individual elements…" only apply to such cases.
In addition, elements can also be released in a reaction so that they would then stand on the right side of the reaction arrow.
Example:
Magnesium reacts with sulfuric acid to form magnesium sulfate and hydrogen.
Mg + H 2 SO 4 → MgSO 4 + H 2
But the rest is true…
LG from the Waterkant
This is a very helpful explanation! 🤩
So I understood it / I want to make sure I understood it correctly:
Normally, the left, that is, in front of the reaction arrow, is always the individual elements, right? And to the right, next to the reaction arrow, the compound formed from it, right?
And in the case of the elements that always stand individually, it must always be taken into account that according to H,O,N,Cl,Br,I,F a profound 2 has to stand, since they occur in nature only 2-atomy.
In the case of connections, however, this rule does not exist.
You have to distinguish between an element and a compound ! An element always contains only one type of atom (Na, S, O, F, etc.). Unless another element symbol comes into play, it's an element. And among elements, H, N, O, F, Cl, Br, and I occur as diatomic minimolecules.
In a compound (ie when different element symbols appear next to each other in the formula), the symbols can be represented in different frequencies.
Example (for oxygen compounds):
Water: H 2 O (1 x O)
Carbon dioxide: CO 2 (2 x O)
Sodium nitrate: NaNO 3 (3 x O)
Potassium permanganate: KMnO 4 (4 x O) 0
Vanadium pentoxide: V 2 O 5 (5 x O)
Glucose: C 6 H 12 O 6 (6 x O)
Potassium dichromate: K 2 Cr 2 O 7 (7 x O)
Octanoctanol : C8H18O8 ( 8xO )
Sodium glucose 6-phosphate: NaC 6 H 11 PO 9 (9 x O)
Phosphorus pentoxide: P 4 O 10 (10 x O)
etc.
As I said, it's important to distinguish between elements and compounds!
Greetings from the waterfront
Thank you very much! Wish you a good improvement! 🍀
I still have a small question, which hopefully can be answered very quickly if you find the pleasure, strength and time:
It is said that H,O,N,Cl,Br,I,F are only 2-atomic in nature and therefore you have also calculated in your answer 19*2 =38, with respect to the fluorine.
In a YouTube video I have just heard that the elements are always written in molecular form even in reaction equations.
Does it mean that always after H,O,N,Cl,Br,I,F a low-level 2 has to stand?
Why is this (excerpt from YouTube video) not like this:
2 H2 + O2 —> 2 H2O
After the arrow, on the right side there is no depth 2 behind the O.
I'm currently very cold so I don't have the power to answer your first question (lasts too long). I have already explained this very often in the forum (representation of a reaction scheme). Do you have to look…
And yes, a factor (number before a formula) always refers to all element symbols (plus indices) in the formula. In contrast, an index (which is the lowest number behind an element symbol or a bracket) refers exclusively to the end standing directly in front of it.
Examples:
2 2 SO 4
is 2 (factor) x 2 (index) Na (= 4 Na)
2 (factor) S (= 2 S)
2 (factor) x 4 (index) O (= 8 O).
3 Ca(NO) 3 ) 2
is 3 (factor) Ca (= 3 Ca)
3 (factor) x 2 (index behind the clamp) N (= 6 N)
3 (factor) x 3 (index on O) x 2 (index behind the clamp) O (= 18 O)
LG from the Waterkant
Ahh, I know. I just watched a video on YouTube and it was said that
HONClBrIF always needs *2. So:
H2, O2, Cl2, Br2, I2, F2.
This was so overlaid that the uncle (chemist) of Jan, Jan sent a letter and he wrote it that way.
This is how I can finally remember it well and have added something important.
ANNEX I have two questions:
In the video were called two reaction equations, a false and a correct one.
The first is wrong, since O can only occur as a two-atom minimolecule.
Yes, a fluoro-ATOM has the mass of 19 u, but the smallest particles of fluoro-gas are not single atoms, but two-atom minimolecules (F 2 ). Thus, two atoms are always present in such a smallest particle of fluorine gas. Therefore it is also (2 • 19 =) 38 u or 38 g/mol…
Where do you come to 38 g/mol? Fluorine has 19.00 u in the periodic system of the elements and this is the same as 19.00 g/mol, right? Maybe you can explain my mistake.
How is the value otherwise calculated?
Yes, it helps if you know the value by heart.
Thank you for this wonderful and very helpful answer to my comment. 🤩
How to calculate the value of 22.4 liters? How do you get that? Should you know that by heart?