Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.
Answer:
7.89 g
Explanation:
Step 1: Write the balanced equation
S₈ + 16 F₂(g) → 8 SF₄
Step 2: Calculate the moles corresponding to 2.34 g of S₈
The molar mass of S₈ is 256.52 g/mol.
Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈
The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol
Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄
The molar mass of SF₄ is 108.07 g/mol.
a) Two or more atoms held together with bonds make up a molecule. b) Pure substances are made of only one type of atom. c) At least two types of atoms are required to make a compound. d) Mixtures can be made of two elements, two compounds, or an element & a compound.
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.
