Answer choice C. Reduction has its electrons as a reactant.
Answer:
Explanation:
We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.
The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.
Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.
To convert to grams, find the molar mass of Ag from the periodic table.
Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.
(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)
Hands off, it's mine.
Answer:
0.04 M
Explanation:
Given data:
Mass of Na₂SO₄= 14.2 g
Volume of solution = 2.50 L
Molarity of solution = ?
Solution:
Number of moles of Na₂SO₄:
Number of moles = mass/ molar mass
Number of moles = 14.2 g/ 142.04 g/mol
Number of moles = 0.1 mol
Molarity :
Molarity = number of moles of solute / volume of solution in L
Molarity = 0.1 mol / 2.50 L
Molarity = 0.04 M
<h3>
Answer:</h3>
150000 J
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Thermodynamics</u>
Specific Heat Formula: q = mcΔT
- <em>q</em> is heat (in J)
- <em>m</em> is mass (in g)
- <em>c</em> is specific heat (in J/g °C)
- ΔT is change in temperature (in °C or K)
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>m</em> = 225 g
[Given] <em>c</em> = 4.184 J/g °C
[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C
[Solve] <em>q</em>
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
- Multiply: q = (941.4 J/°C)(159.8 °C)
- Multiply: q = 150436 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
150436 J ≈ 150000 J
Topic: AP Chemistry
Unit: Thermodynamics
Book: Pearson AP Chemistry
