<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:
This process, which is the opposite of vaporization, is called condensation. As a gas condenses to a liquid, it releases the thermal energy it absorbed to become a gas. During this process, the temperature of the substance does not change. The decrease in energy changes the arrangement of particles.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Explanation:
Note: Molar masses of elements can be found online or in the periodic table.
Moles of Magnesium
= 3.60g / (24.3g/mol) = 0.148mol.
Moles of Chlorine
= 10.65g / (35.45g/mol) = 0.300mol.
Mole ratio of Magnesium to Chlorine
= 0.148mol : 0.300mol = 1 : 2.
Hence we have the empirical formula MgCl2.
Moles of Lithium
= 9.1g / (6.94g/mol) = 1.311mol.
Moles of Oxygen
= 10.4g / (16g/mol) = 0.650mol.
Moles ratio of Lithium to Oxygen
= 1.311mol : 0.650mol = 2 : 1.
Hence we have the empirical formula Li2O.
This is hard to show but here is how you would determine these. NOTE each dot is an electron.
<span>Question 1) </span>
<span>F-H </span>
<span>1) determine the valance electrons for each. F has 7 and H has 1 </span>
<span>2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F </span>
<span>.. </span>
<span>F-H </span>
<span>¨ </span>
<span>Question 2) </span>
<span>2) H-O-H </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place </span>
<span>H-O-H: place two dots above and below the oxygen </span>
<span>Question 3) </span>
<span>3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each </span>
<span>O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place </span>
<span>N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place </span>
<span>place the 2 dots on top and bottom of oxygen. </span>
<span>place 2 above the N </span>
