You can pick it up and move it
They bond because they want to make their outer electron shells more stable
Hope this helps
Have a happy holidays
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer:
1.645 moles of excess reactant that is of magnesium metal are left over.
Explanation:
Moles of magnesium metal = 3.29 mol
Moles of HCl = 3.29 mol
According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :
of magnesium metal
Moles of HCl left = 3.29mol - 3.29 mol = 0
Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol
1.645 moles of excess reactant that is of magnesium metal are left over.
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
