The number of moles of aluminium that are needed to react completely with 13.2 moles of FeO is 8.8 moles
calculation
2Al + 3FeO → 3aFe +Al2O3
by use of of mole ratio of Al: FeO from equation above = 2:3 the moles of Al is therefore
= 13.2 x 2/3=8.8 moles of Al
If a wave was traveling at 15.0 m/s it would take 80 minutes
The empirical formula is SCl_2.
The <em>empirical formula</em> (EF) is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the <em>molar ratio </em>of S to Cl.
Assume that you have 100 g of sample.
Then it contains 31.14 g S and 68.86 g Cl.
<em>Step</em> 1. Calculate the <em>moles of each element</em>
Moles of S = 31.14 g S × (1 mol S/(32.06 g S) = 0.971 30 mol S
Moles of Cl = 68.86 g Cl × (1 mol Cl/35.45 g Cl) = 1.9425 mol Cl
<em>Step 2</em>. Calculate the <em>molar ratio</em> of each element
Divide each number by the smallest number of moles and round off to an integer
S:Cl = 0.971 30: 1.9425 = 1:1.9998 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = SCl_2
Answer:
Yes, Mass is conserved.
Explanation:
Every chemical reactions obey the law of conservation of mass. The law of conservation of mass states that in chemical reactions, mass is always constant.
Equation:
2Na + Cl₂ → 2NaCl
From the equation above, one can observe that the reaction started using 2 atoms of Na and it produced 2 atoms of the same element in NaCl. A molecule of Cl produced 2 atoms of Cl in the NaCl
Design a simple experiment to support your answer:
Aim: To demonstrate the law of conservation of mass
One Na atom weighs 23g
Two Na atom will weigh 2 x 23 = 46g
1 atom of Cl is 35.5g
1 molecule of Cl containing two atoms of Cl will weigh 2 x 35.5 = 71g
Total mass of reactants = mass of 2Na + 1Cl₂ = (46 + 71)g = 117g
On the product side, Mass of 1 NaCl = 23+ 35.5 = 58.5g
Two moles of NaCl will give 2 x 58.5g = 117g
Since the mass on both side is the same, one can say mass is conserved.
<span>Answer:
A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en
so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+
Kf for Cu(en)2^2+ is 1x10^20.
so
1 Cu+2 & 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2
[Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Cu+2 = 9.26 e-19 Molar
since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
