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>
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%.
Answer:
C 2
Explanation:
2Nal + Cl2 → 2NaCl + I2
This is balanced equation
Answer: 1820 mL (to 3 sf)
Explanation:
The atomic mass of aluminum is 26.9815385 g/mol, so 4.10 g of aluminum is equal to 4.10/26.9815385 = 0.15195575300497 moles of Al.
From the coefficients of the equation, we know that for every 2 moles of aluminum consumed, 6 moles of HCl are consumed.
So, this means we need 0.15195575300497(6/2) = 0.45586725901491 moles of HCl.
Substituting into the molarity formula,
- 0.250 = 0.45586725901491/(liters of HCl)
- liters of HCl = 0.45586725901491/0.250
- liters of HCl = 1.8234690360596 L = 1820 mL (to 3 sf)
