<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:
Reducing or increasing the amount of H+ ions / hydronium (H3O+) ions
Explanation:
To reduce the pH (reducing the strength of the acid) can be done by adding a base (including a conjugate base such as bicarbonate ion) which will absorb the H+ ions either through adsorption or reaction.
Adding more H+ decreases the pH of the acid making it stronger. This can be done by adding HCL that will dissociate and increase the H+ ions.
Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:
10 mol NH3 ( 6.022 x 10^23 molecules / 1 mol ) = 6.022x10^24 molecules NH3
NaCl:
Na = + 1
Cl = - 1
hope this helps!
Answer:
34 g/100 mL
Explanation:
The solubility of a compound can be expressed in g/100mL, for this we must divide the mass of the compound that dissolves in the solute by the volume of the solvent.
The solvent, in this case, is water, and that mass of the solute X that dissolved is the mass that was recovered after the solvent was drained and evaporated. So the solubility of X (S) is:
S = 0.17 kg/5L
S = 170g/5000mL
S = 170g/(5*1000)mL
S = 34 g/100 mL
