The correct answer to your question is: <span>C) tin (IV) bromide, SnBr₄</span>
Answer:
- The abundance of 107Ag is 51.5%.
- The abundance of 109Ag is 48.5%.
Explanation:
The <em>average atomic mass</em> of silver can be expressed as:
107.87 = 106.90 * A1 + 108.90 * A2
Where A1 is the abundance of 107Ag and A2 of 109Ag.
Assuming those two isotopes are the only one stables, we can use the equation:
A1 + A2 = 1.0
So now we have a system of two equations with two unknowns, and what's left is algebra.
First we<u> use the second equation to express A1 in terms of A2</u>:
A1 = 1.0 - A2
We <u>replace A1 in the first equation</u>:
107.87 = 106.90 * A1 + 108.90 * A2
107.87 = 106.90 * (1.0-A2) + 108.90 * A2
107.87 = 106.90 - 106.90*A2 + 108.90*A2
107.87 = 106.90 + 2*A2
2*A2 = 0.97
A2 = 0.485
So the abundance of 109Ag is (0.485*100%) 48.5%.
We <u>use the value of A2 to calculate A1 in the second equation</u>:
A1 + A2 = 1.0
A1 + 0.485 = 1.0
A1 = 0.515
So the abundance of 107Ag is 51.5%.
Answer:
A metallic bond.
Explanation:
Potassium is a metal (alkali metal), hence its bonds are metallic bonds.
Hope this helped!
Something like that wouldn't dissolve in oil
<u>Given:</u>
Mass of Na = 115 g
Excess Cl2
<u>To determine:</u>
Mass of NaCl produced
<u>Explanation:</u>
Given reaction is-
2Na(s) + Cl2(g) → 2NaCl(s)
Since Cl2 is in excess, Na will be the limiting reagent
As per the reaction stoichiometry Na:NaCl = 1:1
i.e. moles of Na reacted = moles of NaCl formed
Now, # moles of Na = mass of Na/atomic mass
= 115 g/23 g.mol-1 = 5 moles
Therefore, moles of NaCl = 5
Molar mass of NaCl = 58 g/mol
Mass of NaCl = 5 moles * 58 g.mol-1 = 290 g
Ans: Amount of Nacl produced = 290 g
