<h3>Answer:</h3>
1.47 × 10²¹ molecules OF₂
<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>Atomic Structure</u>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:
</h3>
<u>Step 1: Define</u>
0.132 g OF₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of O - 16.00 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of OF₂ - 16.00 + 2(19.00) = 54.00 g/mol
<u>Step 3: Convert</u>
- Set up:

- Divide/Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.47204 × 10²¹ molecules OF₂ ≈ 1.47 × 10²¹ molecules OF₂
Answer:
water evaportaes from heat it then turns into a gas then can go into a solid one day and repeat the cycle
Explanation:
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
Using a calculator:
(2.568 x 5.8)/4.186 = 3.5581460…
= 3.56 (3sf)
You didn’t specify the correct number of significant figures needed.