Answer:
Approximately 2000 J.
General Formulas and Concepts:
<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)
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>c</em> = 0.897 J/g °C
[Given] <em>m</em> = 79 g
[Given] ΔT = 28°C
[Solve] <em>q</em>
<em />
<u>Step 2: Solve for </u><em><u>q</u></em>
- Substitute in variables [Specific Heat Formula]: q = (79 g)(0.897 J/g °C)(28 °C)
- Multiply [Cancel out units]: q = (70.863 J/°C)(28 °C)
- Multiply [Cancel out units]: q = 1984.16 J
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
1984.16 J ≈ 2000 J
Answer:
20.0 grams
Explanation:
If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.
20.0
*1mL=20.0 grams
They are studying the planet's ecosystem.
<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<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
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
Answer:
Cl2(g) (green/yellow mix) + 2KBr(s) (white) ---> 2KCl(s) (violet) + Br2(g) (reddish brown)
This chemical reaction is a redox type.
Explanation:
Look at the oxidation state, when the number increase your element gets oxidated, when the number decrease, the elements it's getting reduced.