A. NaCl(s) and O2(g)
B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)
C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3
D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)
E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl
F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2
G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2
H. Percent yield = 10/45.1 • 100% = 22.2% yield
<h3>
<u>ANSWER</u></h3>
2. Neon, only
<h3>
<u>EXPLANATION</u></h3>
When Na and F combine they form an electron configuration of 2-8. Na electron configuration is 2-8-1 while F is 2-7, so when they form an ionic bond F will gain Na outermost electron to complete its valence shell due to having a higher negativity. Neon has an electron configuration of 2-8 but argon has 2-8-8.
Answer:
7.8 grams per cm
Explanation:
to get density you need the mass and volume then you divide them so
81.9 grams/10.5 cm gives 7.8g/cm
So the question ask to separate the reaction of the element if your problem into its component half reaction and the best answer or the elements are the following: (Cs-> Cs+ + 1e-) x2 and <span>2e-+Cl2->2Cl-. I hope you are satisfied with my answer and feel free to ask for more </span>
The radius of the anion is 7.413 nm
<h3>How to calculate the force of attraction between charges</h3>
The force of attraction (F) is given by the formula:
- F = (1/4π∈r²)(Zc*e)(Za*e)
where:
∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m
Zc = charge on the cation = +2
Zc = charge on the anion = -2
e = charge on an electron = 1.602 * 10⁻¹⁹ C
r = interionic distance
r = rc + ra
where rc and ra are the radius of the cation and anion respectively
F = 1.64 * 10⁻⁸ N
Therefore based on the equation of force of attraction:
1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²
r² = 5.63 * 10⁻¹⁷
r = 7.50 nm
Since r = rc + ra
where rc = 0.087 nm
thus, ra = r - rc = 7.50 - 0.087
ra = 7.413 nm
Therefore, the radius of the anion is 7.413 nm
Learn more about ionic radius at: brainly.com/question/2279609