Your answer would be 753,000 gallons of water
<h3>
Answer:</h3>
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
Answer: 0.027837938656318
Explanation: Because 1 grams FeO is equal to 0.013918969328159 mole.
The ostrich runs 21 N then goes back by 25 S
The displacement is -4 S
21 - 25 = -4
Answer:
145.78g/mol
Explanation:
Let the time taken for the two gas to effuse be x
For O2:
Volume = 6.51mL
Time = x
Rate (R1) = volume/time
R1 = 6.51/x
For the unknown gas:
Volume = 3.05mL
Time = x
Rate (R2) = Volume/time
R2 = 3.05/x
Now, we can obtain the molar mass of the unknown gas as follows:
R1 = 6.51/x
M1 (O2) = 16 x 2 = 32g/mol
R2 = 3.05/x
M2(unknown gas)
R1/R2 = √(M2/M1)
(6.51/x)/(3.05/x) = √(M2/32)
6.51/3.05 = √(M2/32)
Take the square of both side
(6.51/3.05)^2 = M2/32
Cross multiply to express in linear form:
M2 = (6.51/3.05)^2 x 32
M2 = 145.78
Therefore, the molar mass of the unknown gas is 145.78g/mol