Options? I'm thinking P Waves.
Answer:
1. 7.256g of NaCl
2. 47.33g of Cl2
Explanation:
2 moles of Na reacts to produce 2 moles of NaCl
8 moles of Na will still produce 8 moles of NaCl
Mass of NaCl = molar mass of Nacl/moles of Nacl
=58.5/8
=7.256g of NaCl
From the equation, 2 moles of Na reacts with 1 mole of Cl2
3/2 moles of Cl2 will react with 3 moles of Na
Mass of Cl2 = 71/1.5
=47.33g of Cl2
Answer:
Second one
Explanation:
I think it's the second cause there's no energy w/o heat. So it's the second.
Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
First write all of the compounds/atoms in either side then fill in existing values and balance
Na- 1
Br- 1
Ca- 1
Cl- 2
Na- 1
Cl- 1
Ca-1
Br-2
Balance to get
2NaBr+CaCl2=2NaCl+CaBr2
