Answer is: n (number of fluorine atoms) is 6, formula of the compound is XeF₆.
m(F) = 0.8682 g; mass of fluorine.
n(F) = m(F) ÷ M(F).
n(F) = 0.8682 g ÷ 19 g/mol.
n(F) = 0.0457 mol; amount of substance.
m(Xe) = 1.00 g.
n(Xe) = 1.00 g ÷ 131.293 g/mol.
n(Xe) = 0.00761 mol.
n(Xe) : n(F) = 0.00761 mol : 0.0457 mol.
n(Xe) : n(F) = 1 mol : 6 mol.
<span>Atmospheric Pressure</span>
Answer:
Rubidium and cesium
Explanation:
It is noteworthy to say here that larger cations have more stable superoxides. This goes a long way to show that large cations are stabilized by large cations.
Let us consider the main point of the question. We are told in the question that the reason why potassium reacts with oxygen to form a superoxide is because of its low value of first ionization energy.
The implication of this is that, the other two metals that can be examined to prove this point must have lower first ionization energy than potassium. Potassium has a first ionization energy of 419 KJmol-1, rubidium has a first ionization energy of 403 KJ mol-1 and ceasium has a first ionization energy of 376 KJmol-1.
Hence, if we want to validate the hypothesis that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy, we must also consider the elements rubidium and cesium whose first ionization energies are lower than that of potassium.
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
Answer : The correct option is, (B) 
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of unknown gas = 
= rate of effusion of oxygen gas = 
= molar mass of unknown gas = ?
= molar mass of oxygen gas = 32 g/mole
Now put all the given values in the above formula 1, we get:
The unknown gas could be carbon dioxide 
that has approximately 44 g/mole of molar mass.
Thus, the unknown gas could be carbon dioxide
