Answer:
41.4%
Explanation:
The salt copper II fluoride tetrahydrate is a hydrated salt. A hydrated salt is a salt that has some molecules of water of crystalization attached to its structure. The formula of the salt is CuF2.4H2O
The first step is to calculate the molecular mass of the compound as follows
[64 + 2(19) + 4((2×1) + 16)] = [64 + 38 + 72]= 174gmol-1
Mass of water of crystallization= 4((2×1) + 16) = 72 g
Percentage by mass of water of crystallization= 72 / 174 ×100
Percentage by mass of water if crystalization = 41.4%
Answer:
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized
Answer:
Explanation:
Use Dalton's law and the vapor pressure of water at 23.0 o C to correct the pressure to units of atmoshperes.
PT = Poxygen +Pwater
At 23.0 o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)
762 mmHg = Poxygen + 21.1 mmHg
Poxygen = 762 mmHg - 21.1 mmHg
Poxygen =741 mmHg
Convert the corrected pressure to atmospheres.
(741 mmHg) (1 atm / 760 mmHg) = 0.975 atm
Use the ideal gas law to find out how many moles of gas were produced:
PV = nRT (remember to put volume in liters and temperature in Kelvin)
(0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)
n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)
n = 7.69 X 10-4 mol
Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.
(7.69 X 10-4 mol) (32.0 g / 1 mol) = 2.46 X 10-2 g
The moles of oxygen required to burn Butane is 6 moles.
<h3>What is a Combustion Reaction?</h3>
A reaction in which fuel gets oxidised by an oxidising agent producing a large amount of heat is called a combustion reaction.
In this question
Butane is burnt with oxygen
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol
Balanced equation for the reaction:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Mole ratio C₄H₁₀ : O₂ = 2 : 13
The given mass = 54grams
moles = 54/58 = 0.93 moles
The mole of oxygen required =
0.93/ x = 2/13
0.93*13/2 = x
x = 6.045 moles
Therefore 6 moles of oxygen are required to burn Butane.
To know more about Combustion Reaction
brainly.com/question/12172040
#SPJ1
