The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
Potassium oxide: K₂O.
There's no need for prefixes since K₂O is an ionic compound.
<h3>Explanation</h3>
Find the two elements on a periodic table:
- Potassium- K- on the left end of period four.
- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
- Oxygen is a nonmetal.
A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:
- Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
- Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion
when it combines with metals.
The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each
ion is twice that on a
ion. Each
would pair up with two
. Hence the subscript in the formula:
.
There are two classes of compounds:
- Covalent compounds, which need prefixes, and
- Ionic compounds, which need no prefix.
Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide
, the prefix di- indicates that there are two oxygen atoms in the formula
. However, there's no need for prefix in ionic compounds such as
.
Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.
<h3>
Answer:</h3>
23.459 g NaNO₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄
[Given] 24.14714 g Na₂SO₄
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂
Molar Mass of Na - 22.99 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol
Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We need 5 sig figs (instructed).</em>
23.4587 g NaNO₂ ≈ 23.459 g NaNO₂