Answer:
no .........................
0.008 ÷ 51.3 = 0.0002
Sig Figs
1
0.0002
Decimals
4
0.0002
Scientific Notation
2 × 10-4
E-Notation
2e-4
Words
zero point zero zero zero two
I HOPE I HELP
The answer is D. 459 g As
Answer:
the ion present in the original solution is Ca2+
Explanation:
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate.
<u>Step1</u> : If we add Nacl to the solution, there is no precipitate formed
⇒The only possible ion that can form a precipate with Cl- is Ag+; since there is no precipitate formed, Ag+ is not present
<u>Step2</u> : If we add Na2SO4 to the solution, a white precipitate is formed
The possible ions to bind at SO42- are Ca2+ and Fe2+
But the white precipitate formed, points in the direction of Ca2+
⇒This means calcium is present
<u>Step3</u> : If we add Na2CO3 to the filtered solution, there is a precipate formed
Ca2+ will bind also with CO32- and form a precipitate
So the ion present in the original solution is Ca2+
Answer:
Option B. Cation that is smaller than the original atom.
Explanation:
Magnesium is a divalent element. This implies that magnesium can give up 2 electrons to become an ion (cation) as shown below:
Mg —> Mg²⁺ + 2e¯
Next, we shall write the electronic configuration of magnesium atom (Mg) and magnesium ion (Mg²⁺). This can be written as follow:
Mg (12) = 2, 8, 2
Mg²⁺ (10) = 2, 8
From the above illustration, we can see that the magnesium atom (Mg) has 3 shells while the magnesium ion (Mg²⁺) has 2 shells.
This simply means that the magnesium ion (Mg²⁺) i.e cation is smaller that the original magnesium atom (Mg).
