Answer:
0.095M
Explanation:
HClO4 + NaOH = NaClO4 + H2O
Concentration of acid CA= 0.170M
Concentration of base CB= ???
Volume of base VB= 28.5ml
Volume of acid VA= 16.0ml
Number of moles of acid nA= 1
Number of moles of base nB= 1
From
CAVA/CBVB= nA/nB
CB= CAVAnB/VBnA
CB= 0.170×16.0×1/28.5×1
CB= 0.095M
6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe
<h3>Explanation</h3>
Method One: Refer to electron transfers.
Oxidation states:
- Na: from 0 to +1; loses one electron.
- Fe: from +3 to 0; gains three electrons.
Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.
6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe
- There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
- There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Method Two: Atoms conserve.
Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.
? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe
There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.
? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.
<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Answer:
a) 0,5
Explanation:
If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5
Answer:
The mass fraction of ferric oxide in the original sample :
Explanation:
Mass of the mixture = 3.110 g
Mass of 
Mass of 
After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.
Mass of mixture left after all the ferric oxide has reacted = 2.387 g
Mass of mixture left after all the ferric oxide has reacted = y
The mass fraction of ferric oxide in the original sample :
