Answer:
large supply of nutrients.
Explanation:
In a wetland, the soil is covered by water or is almost covered by water. This water may be coming upwards from an underground aquifer. Wetlands are covered by water for most of the year.
They are sheltered waters and do provide habitats for many living things.
Nutrients such as; Carbon sulfur, phosphorus, carbon, and nitrogen are cycled within the soil of wetlands hence wetlands have a large supply of nutrients.
<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer:
0.733 mol.
Explanation:
- From the balanced equation:
<em>2Fe₂O₃ + C → Fe + 3CO₂,</em>
It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.
- Since Fe₂O₃ is in excess, C will be the limiting reactant.
<u><em>Using cross multiplication:</em></u>
1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.
??? mole of C produces → 2.2 moles of CO₂.
∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.
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
