Answer:
0.324 g is required to make 5.00 M solution of NaCl in 0.800 L.
Given data:
Molarity = 5.00 M
Formula Mass = 58.5 g/mol
Required volume = 0.800 L
To Find;
Mass in gram = ?
Solution:
Formula for calculating mass in gram is given as,
Mass in gram = Molarity × Formula mass × Volume required / 1000 putting values
Mass in gram = 5.00 M × 58.5 g/mol × 0.800 L / 1000
Mass in gram = 0.234 g
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Answer:
0.6 moles of CaO will produced.
Explanation:
Given data:
Mass of calcium = 23.9 g
Moles of CaO produced = ?
Solution:
Chemical equation:
2Ca + O₂ → 2CaO
Number of moles of calcium:
Number of moles = mass/ molar mass
Number of moles = 23.9 g / 40 g/mol
Number of moles = 0.6 mol
Now we will compare the moles of calcium and CaO.
Ca : CaO
2 : 2
0.6 : 0.6
0.6 moles of CaO will produced.
Should be , b
. Not positive tho
