Answer:
1.69 g Na₂O
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.0273 mol Na₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of Na - 22.99 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Na₂O - 2(22.99) + 16.00 = 61.98 g/mol
<u>Step 3: Convert</u>
<u />
= 1.69205 g Na₂O
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
1.69205 g Na₂O ≈ 1.69 g Na₂O
Explanation: In IUPAC, E-Z convention is given for describing the cis - trans notation to the isomers. According to CIP rule, the groups on the doubly bonded carbon atoms are given priorities based on the the atomic masses of first connected atom.
If the highest priority groups are on the same side, it is known as Z-form and if the highest priority groups are on opposite side, it is known as E-form.
We are given (Z)-3-bromo-6-methyl-2-heptene, in this the highest priority groups are bromine on one side and methyl- group on another side.
The structure is provided in the image below.
The substance will not sink in water, as it Dencity is (0.5)is less than the density of water (1gcm).
Explanation:
Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.
C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)
When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.
When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.
Answer:
D ≈ 4.86 g/mL
Explanation:
Density = Mass over Volume
D = m/V
Step 1: Define variables
m = 13.76 g
V = 2.83 mL
D = unknown
Step 2: Substitute and Evaluate for Density
D = 13.76 g/2.83 mL
D = 4.8621908127208480565371024734982 g/mL
Step 3: Simplify
We are given 3 sig figs.
4.8621908127208480565371024734982 g/mL ≈ 4.86 g/mL
