Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt
= 11.75 g - 9.25 g = 2.50 g
number of water moles = 2.50 g / 18 g/mol = 0.139 mol
number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol
ratio of water moles to CoCl₂ moles - 0.139 mol / 0.0712 mol = 1.95
rounded off 2 moles of water for every 1 mol of CoCl₂
formula - CoCl₂.2H₂O
name - Cobalt(II) chloride dihydrate
Potassium 23.5g/39.0983g/mol = 0.601mol
The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol
Therefore 0.3005mol of F2 is needed to find liters use
formula V = nRT/P (V)Volume = 22.41L
(T)Temperature = 273K or 0.0 Celsius
(P)Pressure = 1.0atm
<span>(R)value is always .08206 with atm n = 0.3005moles
(273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>
Using the significant figure it would be 27.3
Answer:
3.81 g Pb
Explanation:
When a lead acid car battery is recharged, the following half-reactions take place:
Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)
Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻
We can establish the following relations:
- 1 A = 1 c/s
- 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
- The molar mass of Pb is 207.2 g/mol
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)
Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:
