Answer:
1.79 mol.
Explanation:
- For the balanced reaction:
<em>2NaCl + F₂ → 2NaF + Cl₂.
</em>
It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.
- Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = 18.3 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (299 K).
∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.
- Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:
<em><u>Using cross multiplication:</u></em>
2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.
??? mol of NaCl is needed to react with → 0.895 mol of F₂.
∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.
An orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy.
Explanation:
The only true statement from the given options is that "an orbital that penetrates into the region occupied by core electrons is less shielded from nuclear charge than an orbital that does not penetrate and therefore has a lower energy." Inner orbitals which are also known to contain core electrons feels the bulk of the nuclear pull on them compared to the outermost orbitals containing the valence electrons.
- The nuclear pull is the effect of the nucleus pulling and attracting the electrons in orbitals.
- This pull is stronger for inner orbitals and weak on the outer ones.
- The outer orbitals are said to be well shielded from the pull of the nuclear charge.
- Also, based on the quantum theory, electrons in the outer orbitals have higher energies because they occupy orbitals at having higher energy value.
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Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer: 1.24 × 10^25
Explanation:
× 
Using our knowledge in unit conversions, we know the mole units cancel each other out and all there's left is the atom unit. From here we can multiply the fractions and eventually we end with the number 124.0532 × 10^23
According to the scientific notation rules, the number to the left of the decimal cannot exceed 10 so we have to move the decimal to the left two spaces. With this change, we also have to change the exponent of the 10. Because we moved the decimal point two spaces to the left, that means we have 10^25.
