Answer:
The answer to your question is V2 = 825.5 ml
Explanation:
Data
Volume 1 = 750 ml
Temperature 1 = 25°C
Volume 2= ?
Temperature 2 = 55°C
Process
Use the Charles' law to solve this problem
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 25 + 273 = 298°K
T2 = 55 + 273 = 328°K
-Substitution
V2 = (750 x 328) / 298
-Simplification
V2 = 246000 / 298
-Result
V2 = 825.5 ml
Answer:
<em>They could see how the spaceship is going to launch and how the trip is going to end up. I hope this helped...!</em>
Explanation:
<u>Chemistry</u> can be used for lots of things, for example, <em>engineering</em> and <em>science</em> such as <em>kinetic energy</em> or <em>mechanical energy</em>.
Answer:
To the right
Explanation:
Step 1: Given data
- Partial pressure of PCl₅ (pPCl₅) = 0.548 atm
- Partial pressure of PCl₃ (pCl₃) = 0.780 atm
- Partial pressure of Cl₂ (pCl₂) = 0.780 atm
Step 2: Write the balanced equation
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Step 3: Calculate the pressure reaction quotient

Step 4: Determine whether the reaction proceeds to the right or to the left as equilibrium is approached
Since <em>Qp < Kp</em>, the reaction will proceed to the right to attain the equilibrium.
Answer:
1.21 mol KClO₃
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Mole Ratio
<u>Stoichiometry</u>
- Analyzing reactions rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[rxn] 2KClO₃ → 2KCl + 3O₂
[Given] 58.3 g O₂
[Solve] mol KClO₃
<u>Step 2: Identify Conversions</u>
[rxn] 2 mol KClO₃ → 3 mol O₂
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of O₂: 2(16.00) = 32.00 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.21458 mol KClO₃ ≈ 1.21 mol KClO₃