Answer:
option C is correct = 1.14 × 10²² molecules of CO₂
Explanation:
Given data:
Number of moles of CO₂ = 0.0189 mol
Number of molecules = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
For given question:
1 mole of CO₂ = 6.022 × 10²³ molecules of CO₂
0.0189 mol of CO₂ × 6.022 × 10²³ molecules of CO₂ / 1mol
1.14 × 10²² molecules of CO₂
Thus, option C is correct.
Answer:
Here are a few more examples:
Smoke and fog (Smog)
Dirt and water (Mud)
Sand, water and gravel (Cement)
Water and salt (Sea water)
Potassium nitrate, sulfur, and carbon (Gunpowder)
Oxygen and water (Sea foam)
Petroleum, hydrocarbons, and fuel additives (Gasoline)
Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.
Examples of Heterogeneous mixtures – air, oil, and water, etc.
Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.
Explanation:
Answer:
6.82 g H₂S
General Formulas and Concepts:
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.200 mol H₂S
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
6.818 g H₂S ≈ 6.82 g H₂S
Full question:
The IUPAC name for CH3CH2C≡CCH3 is:
Answer:
2-pentyne
Explanation:
To name hydrocarbons, you first you have to identify the longest carbon chain. There are 5 carbons in this chain, so we know the name is "pent".
You then have to identify the presence of any double or triple bonds. If double bonds, it is an alkene, if triple bonds, it is an alkyne. In this case there is a triple bond, so we know the hydrocarbon is pentyne.
You then number the chain to give the lowest number to the triple bond. It could either be 4 (countnig carbons from left to right) or 2 (from right to left). Therefore, the answer is 2-pentyne.
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
