Answer:
1019.27 g.
Explanation:
- For the balanced reaction:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear 2 moles of Na with 1 mole of Cl₂ to produce 2 moles NaCl.
- Firstly, we need to calculate the no. of moles of Cl₂ is needed to react with 57.5 mol Na:
2 moles of Na need → 1 mol of Cl₂, from the stichiometry.
57.5 moles of Na need → ??? mol of Cl₂.
<em>∴ The no. of moles of Cl₂ is needed to react with 57.5 mol Na =</em> (1 mol)(57.5 mol)/((2 mol) <em>= 28.75 mol.</em>
<em>∴ the mass of Cl₂ is needed to react with 57.5 mol Na = (no. of moles of Cl₂)(molar mass of Cl₂) =</em> (28.75 mol)(35.453 g/mol) <em>= 1019.27 g.</em>
Answer:
47.5 mL
Solving:
M1 = 4.00 M
V1 = ?
M2 = 0.760 M
V2 = 0.250 L
---
M1 * V1 = M2 * V2
V1 = ( M2 * V2 ) / M1
V1 = ( 0.760 * 0.250 ) / 4.00
V1 = ( 0.190 ) / 4.00
V1 = 0.0475 L
Answer: -
6 mg
Explanation: -
Initial amount N₀ = 192 mg
Half life = 6 hours.
Time given = 30 hours
Number of half lives =
= 5
Amount present after 30 hours = N₀ x []
Amount present after 30 hours = 192 mg x []
= 6 mg