Answer:
8.4 m/s
Explanation:
Initial velocity of the object is 0 as it was at rest.
We need to find the object’s average velocity during the 3 second to 5 second.
We can see from the graph that from 3 s to 5 s, the graph is a straight line. If we take the slope we can find the average velocity during the 3 second to 5 second.
Displacement between 3 s to 5 s is :
D = 24.8 m - 8 m = 16.8 m
Time = 5 s - 3 s = 2 s
Now,
Average velocity,
So, the object’s average velocity during the 3 second to 5 second is 8.4 m/s.
Answer:
New Gas volume (V2) = 5.86 L
Explanation:
Given:
Gas volume (V1) = 2.93 L
Temperature (T1) = 20°Celsius
After Increasing temperature:
New Temperature (T2) = 40°Celsius
Find:
New Gas volume (V2) = ?
Computation:
According to Charles' gas Law:
⇒ Gas volume (V1) / Temperature (T1) = New Gas volume (V2) / New Temperature (T2)
⇒ 2.93 L / 20°Celsius = New Gas volume (V2) / 40°Celsius
⇒ New Gas volume (V2) = 5.86 L
<h3>
Answer:</h3>
690 g AgCl
<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
- Writing Compounds
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Limiting Reactant/Excess Reactant
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂
↓
[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂
[Given] 2.4 mol ZnCl₂
[Solve] <em>x</em> g AgCl
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol ZnCl₂ → 2 mol AgCl
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
<u>Step 3: Stoich</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>
687.936 g AgCl ≈ 690 g AgCl
