Answer:
molarity of diluted solution = 1.25 M
Explanation:
Using,
C1V1 (Stock solution) = C2V2 (dilute solution)
given that
C1 = 2.50M
V1 = 250ML
C2 = ?
V2 = 500ML
2.50 M x 250 mL = C2 x 500 mL
C2 = (2.50 M x 250 mL) / 500 mL
C2 = 1.25 M
Hence, molarity of diluted solution = 1.25 M
Noble gases.
group 18. elements that are all unreactive .
fixed naming!
Answer:
-<em>9</em><em>.</em><em>6</em><em>7</em><em>5</em>
Explanation:
<em>c</em><em>o</em><em>r</em><em>r</em><em>e</em><em>c</em><em>t</em><em> </em><em>m</em><em>e</em><em> </em><em>i</em><em>f</em><em> </em><em>i</em><em>m</em><em> </em><em>w</em><em>r</em><em>o</em><em>n</em><em>g</em><em>.</em><em>!</em><em>!</em><em> </em><em />
<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
Answer:
B. A body at rest will stay at rest until it is acted upon by another object
Explanation:
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
