First solve the moles of oxgen present in the compound
mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H
then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2
Answer:
Percent error = 12.5%
Explanation:
In a measurement you can find percent error following the formula:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.
Replacing:
Percent error = |Measured value - Accepted Value| / Acepted value * 100
Percent error = |19.6L - 22.4L| / 22.4L * 100
Percent error = |-2.8L| / 22.4L * 100
Percent error = 2.8L / 22.4L * 100
Percent error = 12.5%
<h3>
Answer:</h3>
28 mol CaF
<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>
- 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>
[Given] 1.7 × 10²⁵ molecules CaF
[Solve] moles CaF
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
28.2298 mol CaF ≈ 28 mol CaF
Answer:
Explanation:
To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.
The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.
This can be expressed below:
RAM = Σmₙαₙ
where mₙ is the mass of isotope n
αₙ is the abundance of isotope n
for this problem:
RAM of Li = m₆α₆ + m₇α₇
m₆ is mass of isotope Li-6
α₆ is the abundance of isotope Li-6
m₇ is mass of isotope Li-7
α₇ is the abundance of isotope Li-7
