With the equation <span>A₂+2B₂→2AB₂ we know that one mole of A₂ is necessary to react with two moles of B₂ and with that, we obtain two moles of the product. ( you can see this by just seeing what numbers are before the letters)
Since it's needed more quantity of B than A, B is the limited reagent.
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<h2>it is not dirty so do try it >:((((((</h2>
Answer:
pH = 1.6
Explanation:
0 mL NaOH:
⇒ [ H3O+] = M HBr = 0.1 M
⇒pH = -log [H3O+] = 1
30 mL NaOH:
⇒ mol NaOH = 0.1 mol / L * 0.03 L = 3 E-3 mol
⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol
⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M
⇒ pH = - log (0.025) = 1.6
88g ( so sorry if this isn’t correct )
Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
