The molecular formula for a monocyclic hydrocarbon with 14 carbons and 2 triple bond is C₁₄H₂₀
<h3>Molecular formula</h3>
A formula that gives the number of atom of each element present in a one molecule or a compound.
<h3>Monocyclic hydrocarbons</h3>
The name of the saturated hydrocarbons formed by the name attaching the perfix cyclo to the name of acyclic unstaturated hydrocarbon
The molecular formula for a monocyclic hydrocarbon with 14 carbon and 2 triple bonds is C₁₄H₂₀
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The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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Answer:
a)
b)
Explanation:
a) The reaction:
The free-energy expression:
The element wich is reduced is the Fe and the one that oxidates is the Mg:
The electrons transfered (n) in this reaction are 2, so:
b) If you have values of enthalpy and enthropy you can calculate the free-energy by:
with T in Kelvin
Answer:
A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J
B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.
C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero
D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero
E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero
F) if there is no work done, and no heat added, then the internal energy will be equal zero.
Answer : The enthalpy of the reaction is, -2552 kJ/mole
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given enthalpy of reaction is,
The intermediate balanced chemical reactions are:
(1)
(2)
(3)
(4)
Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :
(when we are reversing the reaction then the sign of the enthalpy change will be change.)
The expression for enthalpy of the reaction will be,
Therefore, the enthalpy of the reaction is, -2552 kJ/mole