Answer:
First, find out how many moles of N2I6 you have. Then convert that to grams.
molar mass N2I6 = 789 g
moles N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
grams N2I6 = 0.136 moles x 789 g/mole = 107 g = 110 g (to 2 significant figures)
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Answer: D
Explanation:
the atom of the metal loses one electron which becomes delocalised and is attragted by the positive nucleus leading to formation of metallic bond.
Answer:
4NH₃(g) +3O₂(g) ⇒2N₂(g) +6H₂O(g)
Explanation:
