<h3>
Answer:</h3>
7.4797 g AlF₃
<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] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol KF = 2 mol AlF₃
Molar Mass of K - 39.10 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃
The balanced chemical reaction is written as:
<span>CH4 (g) + 2 O2 (g) ----> CO2 (g) + 2 H2O (g)
</span>
We are given the amount of water to be produced from the reaction. This amount will be used for the calculations. Calculations are as follows:
12.4 L H2O ( 1 mol / 22.4 L ) ( 1 mol CH4 / 2 mol H2O ) ( 22.4 L / 1 mol ) = 6.2 L CH4
Answer:
C20 H14 O2
Explanation:
Remark
This is a sample, which the question does not say and should. It is a fraction of 1 mole. So what you have to do is multiply the numbers given by x and equate it to 286.28
Equation
150,86* x + 8.86*x + 20.1*x = 286.28
179.8x = 286.28
x = 286.26/179.8
x = 1.592
Now multiply the given numbers by 1.592
150.86 * 1.592 = 240.58
8.85 * 1.592 = 14.1
20.1 * 1.592 = 32
Rounding you get
240/12 = 20
14.1/1 = 14
32/16 = 2
C20 H14 O2
Answer:
CH₃CH(CH₃)CH(C₃H₇)CH₂CH(CH₃)₂:
4-isopropyl-2-methylpentane.
Explanation:
Step One: Draw the structure formula of this compound. Parentheses in the formula indicate substitute groups that are connected to the carbon atom to the left.
For example, the first (CH₃) indicates that the second carbon atom from the left is connected to:
- the CH₃- on the left-hand side,
- the -CH(C₃H₇)CH₂CH(CH₃)₂ on the right-hand side,
- a hydrogen atom, and
- an additional CH₃- group that replaced one hydrogen atom.
Each carbon atom in this compound is connected to four other atoms. All bonds between carbon atoms are single bonds.
The C₃H₇ in the second pair of parentheses is the condensed form of CH₃CH₂CH₂-. See the first sketch attached. Groups in parentheses are highlighted.
Step Two: Find the carbon backbone. The backbone of a hydrocarbon is the longest chain of carbon atoms that runs through the compound. See the second sketch attached. The backbone of this compound consists of seven carbon atoms and is highlighted in green. The name for this backbone shall be heptane.
Step Three: Identify and name the substitute groups.
The two substitute groups are circled in blue in the second sketch.
- The one on the right -CH₃ is a methyl group.
- The one on the left is branched.
![\begin{aligned}\text{CH}_3-&\text{CH}-\text{CH}_3\\[-0.5em]&\;|\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctext%7BCH%7D_3-%26%5Ctext%7BCH%7D-%5Ctext%7BCH%7D_3%5C%5C%5B-0.5em%5D%26%5C%3B%7C%5Cend%7Baligned%7D)
This group can be formed by removing one hydrogen from the central carbon atom in propane. The name for this group is isopropyl.
Step Four: Number the atoms.
Isopropyl shall be placed before methyl. Start from the right end to minimize the index number on all substitute groups. The methyl group is on carbon number two and the isopropyl group on carbon number four. Hence the name:
4-isopropyl-2-methylheptane.
