Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water
I hope it helped you!
Red giant would be the answer here
<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>
<h3>
Answer:</h3>
1100 mmHg
<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>Gas Laws</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.5 atm
[Solve] mmHg
<u>Step 2: Identify Conversions</u>
1 atm = 760 mmHg
<u>Step 3: Convert</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>
1140 mmHg ≈ 1100 mmHg
Answer:
=759.95 grams.
Explanation:
The molar mass of chromium is 51.9961 g/mol
Therefore the number of moles of chromium in 156 grams is:
Number of moles =mass/RAM
=156g/51.9961g/mol
=3 moles.
From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.
Therefore 3 moles of chromium produce:
(3×2)/4 moles =1.5 moles of chromium oxide.
I mole of chromium oxide has a mass of 151.99 g
Thus 1.5 moles= 1.5mole ×151.99 g/mol
=759.95 grams.
