First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
What volume of 2.50 m lead ll nitrato
5C2O4^(2-)(aq) + 2MnO4^-(aq) + 16H+(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l)` is the chemical reaction and mole ratio between oxalate and permanganate in the titration reaction.
A chemical reaction is a procedure that causes one group of chemical components to change chemically into another. Chemical reactions, which can frequently be described by a chemical equation, traditionally include changes that only affect the locations of electrons in the formation and dissolution of chemical bonds between atoms, with no change to the nuclei (no change to the elements present). The study of chemical processes involving unstable and radioactive elements, where both electronic and nuclear changes may take place, is known as nuclear chemistry.
To know more about chemical reaction, click here,
brainly.com/question/11231920
#SPJ4
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = =74.8 %.
Answer:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Explanation:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Magnesium is more electro positive than aluminum hence it functions as the anode. Six electrons are lost/gained in the redox process as shown in the oxidation and reduction half reaction equations above. Magnesium is oxidized to magnesium ion while aluminum is reduced to elemental aluminum.