<h3>
Answer:</h3>
8.01 mol MgO
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + O₂ → MgO
[RxN - Balanced] 2Mg + O₂ → 2MgO
[Given] 8.01 moles Mg
[Solve] moles MgO
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Mg → 2 mol MgO
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.
It is called convection. When warm air, or current, moves up and disperse outwards as cold air, or current, moves into the warmer region.
Answer:
The law of conservation of mass represents a balanced chemical equation.
Answer:
46.761g/mol
Explanation:
Given parameters:
Element = Hilarium , Hi
Isotopes: Hi- 45, Hi-46 and Hi- 48
Natural abundance of Hi-45 = 18.3%
Hi-46 = 34.5%
Hi-48 = 47.2%
Unknown:
Atomic weight of naturally occurring Hilarium = ?
Solution:
Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.
We can use the expression below to find this value:
Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈
m is the atomic mass of each isotope and α is the abundance
Atomic weight = (45 x 
) + (46 x 
) + (48 x 
)
Atomic weight of Hi = 8.235 + 15.870 + 22.656 = 46.761g/mol
