Answer:
[A]²
Explanation:
Since the formation is independent of D, D is 0 order.
Since a quadruples when it is doubled it can be written as
2A^X= 4
To find the unknown power we can assume A= 1 to make the math simple. So When a = 2 (Because you doubled it) raised to X power it will equal 4
so the unknown power is 2
Making the rate law
[a]²[b]⁰
or simply just
[A]²
<h3>Answer:</h3>
64 g O₂
<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
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Divide/Multiply [Cancel Units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
Answer:
116.5 g of SO₂ are formed
Explanation:
The reaction is:
S₈(g) + 8O₂(g) → 8SO₂ (g)
Let's identify the moles of sulfur vapor, by the Ideal Gases Law
We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K
5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K
(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈
Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide
Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 = 1.82 moles
We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g
It would be the same thing if the Co does not have a number with it because it can’t reduce