Answer:
A.
Explanation:
While solar power does produce carbon dioxide emissions in the manufacturing processes required to produce panels and batteries, it does not produce CO2 while converting solar energy to electrical energy.
<span>Ionic bonding between sodium and phosphate ions.</span>
Answer: Tall fescue can be harmful to grazing cattle and horses because the grass can become tough and infected with endophytes, causing poor grazing. Switchgrass and tall fescue are less likely to be poisonous to dogs, cats or humans than to horses or cattle, but eating either of them might cause stomach upset.
Explanation:
Answer:
that's because....
group 1 (e.g Na, K) those tend to lose one electron to gain noble gas electron configuration.
they can achieve that by just losing one electron from their outer shell.
as you go down the group 1, element gets bigger in size, which means there is more space between nucleus (which is in center of atom) and electron of outer shell. the more far away they are the less attraction force between them.
so its easier for potassuim to lose one electron than for lithuim.
so that means potassium will easily give up 1 electron to react with non metal or other element therefore it is more reactive than lithuim
but in case of non metal, the opposite happens but simple to understand.
as you go down the group 7 (halogen- Cl, Br, I) element will get bigger therefore force between nucleus and outer electron is getting smaller. they have to gain 1 electron in order to fill the outer shell (to gain noble gas electron configuration.)
as florine is more smaller in size than clorine it is more reactive because florine has more tendency to pull extra electron from metal or other element towards its side. so it easily gain 1 electron to react.
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
