NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27%
3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
0.8211 grams Na + 1.266 grams Cl = 2.087 grams
<span>34.2 grams
Lookup the atomic weights of the involved elements
Atomic weight potassium = 39.0983
Atomic weight Chlorine = 35.453
Atomic weight Oxygen = 15.999
Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol
Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol
The balanced equation for heating KClO3 is
2 KClO3 = 2 KCl + 3 O2
So 2 moles of KClO3 will break down into 3 moles of oxygen molecules.
0.713188188 mol / 2 * 3 = 1.069782282 mols
So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be
1.069782282 * 2 * 15.999 = 34.23089345 grams
Rounding the results to 3 significant figures gives 34.2 grams</span>
A. The molecules of solids are close together and compact, liquids are spread out but not too far apart, and gas molecules are really far apart.
B. Increase in temperature causes pressure to go up. Decrease in temperature cause pressure to go down
Answer:
g NaCl = 424.623 g
Explanation:
<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent
∴ solvent: H2O
∴ mass H2O = 2.314 Kg
mol NaCl:
⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl
∴ mm NaCl = 58.44 g/mol
⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:

The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction
= 0.8325;
If we want to find:

Then:


Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.