Molten barium chloride is separeted
into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
<span>The anode is positive and the cathode is negative.</span>
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
Time, Ice size, water temp
1) List the reactants: sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇).
Reactants undergo change during a chemical reaction.
2) List the products: water (H₂O), carbon dioxide (CO₂) and sodium citrate (Na₃C₆H₅O₇).
Products are the substances formed from chemical reactions.
3) The balanced chemical equation:
3NaHCO₃ + H₃C₆H₅O₇ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇.
We want to solve Q = mcΔT for the liquid water; its change in temperature will tell us the amount of thermal energy that flowed out of the reaction. The specific heat, c, of water is 4.184 J/g °C.
Q = (72.0 g)(4.184 J/g °C)(100 °C - 25 °C) = 22593.6 J
Q ≈ 2.26 × 10⁴ J or 22.6 kJ (three significant figures).
