289.4 F is the right one man
Answer:
m = 65.637 g
Explanation:
Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)
Initial Temperature = 30.°C
Final Temperature = 51°C
Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C
Mass = ?
The relationship between these quantities is given by the equation;
H = mCΔT
where c = 0.444 J/g°C
Inserting the values in the equation;
612 = m * 0.444 * 21
m = 612 / (0.444 * 21)
m = 65.637 g
Answer:- The formula 
tells us that one formula unit of this compound is composed of one calcium atom, two nitrogen atoms, and six oxygen atoms.
Explanations:- Subscripts tell us about the number of atoms of the element for which they are used. For example, here the subscript of Ca is one, it means there is one calcium atom in the given one formula unit.
When we have subscripts inside and outside the parenthesis then they are multiplied and the outside subscript is considered for all the atoms present inside the parenthesis.
Here, for the given chemical formula, the subscript of N is 1 and the subscript present outside is 2. So, 1 x 2 = 2 and for oxygen, 3 x 2 = 6
So, we have one calcium atom, two nitrogen atoms and six oxygen atoms for one formula unit of given compound.
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
