Answer:
I don’t know this answer sorry
Explanation:
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
When connectors are marked with a combination of metals,
it can be used as a connector of one of the metals or an alloy of the two
metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and
Cu is copper, therefore the answer is:
<span>Yes, it is suitable for use with copper, copper-clad
aluminum, and aluminum conductors.</span>
Use a strip of paper covered in PH indicating dye.
