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: C. HNO3 and NaOH
Explanation:
Arrhenius acid produces hydogen ion (H+) when dissolved in water. An example of Arrhenius acid is HNO3.
Arrhenius base produces hydroxyl ion (OH-) when dissolved water. An example of Arrhenius base is NaOH.
2Mg+O₂⇒2MgO
It can be a combustion reaction, or it can be a combination reaction
Answer:
= 0.014 g of BaCO3
Explanation:
Let x = mol/L of BaCO3 that dissolve.
This will give;
x mol/L Ba2+ and x mol/L CO32-
But;
Ksp = 5.1x10^-9.
Therefore;
Ksp = 5.1 x 10^-9 = (x)(x)
Thus;
x = molar solubility
= √ (5.1 x 10^-9)
= 7.1 x 10^-5 M
Therefore;
Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol
= 0.014 g
I think it would be d. all of the above
