Answer : The volume of oxygen at STP is 112.0665 L
Solution : Given,
The number of moles of 
= 5 moles
At STP, the temperature is 273 K and pressure is 1 atm.
Using ideal gas law equation :
where,
P = pressure of gas
V = volume of gas
n = the number of moles
T = temperature of gas
R = gas constant = 0.0821 L atm/mole K (Given)
By rearranging the above ideal gas law equation, we get
Now put all the given values in this expression, we get the value of volume.
Therefore, the volume of oxygen at STP is 112.0665 L
Answer:
N2O2(g) +O2(g) ===> 2NO2(g)
Explanation:
For a nonelementary reaction, the reaction equation is described as the sum of all the steps involved. All these steps constitute the reaction mechanism. Each step in the mechanism is an elementary reaction. The rate law of the overall reaction involves the rate determining step (slowest step) in the reaction sequence.
Now look at the overall reaction 2NO(g) + O2(g) ---------> 2NO2(g)
The two steps in the mechanism are
2NO(g) --------->N2O2(g) (fast)
N2O2(g) +O2(g) -------> 2NO2(g) (slow)
Summing all the steps and cancelling out the intermediate N2O2(g), we obtain the reaction equation;
2NO(g) + O2(g) ---------> 2NO2(g)
Hence the answer.
Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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 Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
- [DA} Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Answer:
49.35 mL
Explanation:
Given: 56.2 mL of gas
To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg
Solution:
At 820 mm of Hg, volume of gas is 56.2 mL
At 1 mm of Hg, volume of gas is 
At 720 mm of Hg, volume of gas is 
