B
Orbitals fill in a way that all of them first gain one electron, and then they start receiving the second electron if any are left. Thus, the three will fill with one electron each first and then the fourth will go into the first.
Molar Mass of KCLO3
= 39 + 35.5 + 48 g
= 122.5 g .
So , percentage of Potassium ( K )
= 39 / 122.5 * 100 %
= 31.83 %
Now , percentage of Chlorine ( Cl )
= 35.5 / 122.5 * 100 %
= 28.97 %
So, finally, percentage of Oxygen ( O )
= 48 / 122.5 * 100 %
= 39.18 %
Solution with a ph of 7 Is neutral
No normal but not unheard of. And I think she might be lying to you about the father if you have only exchanged saliva…
Answer:
1). 1 mole of Carbon burnt in air
C + O2 →CO2
1 mole of carbon produces 1 mole of CO2 which is 44g of CO2
2). 1 mole of carbon is burnt in 16g of dioxygen
32g of O2 = 44g of CO2
1g of O2 = 44/32
CO2 (Dioxygen is limiting reagent)
16g of O2 = 4/32 × 16 = 22g of CO2 in one mole
3) 2 moles of Carbon burnt in 16g of dioxygen
16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2