Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂
<h3>
Answer:</h3>
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
<h3>
Explanation:</h3>
We are given the Equation;
CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl
Assuming the question requires us to balance the equation;
- A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
- Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
- According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
- Balancing chemical equations involves putting appropriate coefficients on the reactants and products.
In this case;
- To balance the equation we are going to put the coefficients 3, 2, 1, and 6.
- Therefore; the balanced equation will be;
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
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