The metalloid that has three valence electrons is Boron~
0.20 moles of iron will be formed in the reaction.
Explanation:
The balanced chemical equation for the reaction between iron (iii) oxide and carbon monoxide to form Fe is to be known first.
the balanced reaction is :
Fe2O3 + 3CO⇒ 2 Fe + 3 CO2
so from the data given the number of moles of carbon monoxide can be known:
3 moles of CO reacted with Fe2O3 to form 2 moles of iron in the reaction.
Number of moles of CO is 6.20 moles
11.6 gm of iron is formed
so the number of moles of iron formed is calculated as
n = mass of iron ÷ atomic weight of iron
= 11.6 ÷ 55.84
= 0.20 moles of iron will be formed when 11.6 gram of iron is produced.
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
Answer:
The substance that remained on the filter paper is Al(OH).
Explanation:
- Filter paper is the substance that is used in laboratory to separate the solid objects. It doesn't filter aqueous and gaseous products.
- So in our experiment the product formed is aluminum hydroxide Al(OH) and sodium chloride (NaCl).
- Between two products, sodium chloride is in aqueous form as indicated in the question. So it wont remain in the filter paper.
- Hence aluminum hydroxide being only solid product remains on the filter paper.
