The standard enthalpy of formation (Δ
) is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements.
Standard enthalpies (Δ) of formation for given reaction is 978.3 kJ
<h3>
What is Standard enthalpies of formation?</h3>
The standard enthalpy of formation is defined as the enthalpy change when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.
Given reaction ;
To Find : Δ
Δ = ∑np Δ (products) – ∑np Δ (reactants)
Δ = [1(Δ 
) + 4(Δ 
)] – [1(Δ 
) + 4(Δ 
)]
Δ = [1(0) + 4(-241.8)] – [1(+9.16) + 4(0)]
Δ = [4(-241.8)] – [1(+9.16)] = 978.3 kJ
Learn more about Enthalpy here ;
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1.661x10 negative 24 squared
Answer:
D. 
Explanation:
Your unbalanced nuclear equation is
It is convenient to replace the question mark by an atomic symbol, 
, where <em>x</em> = the atomic number, <em>y</em> = the mass number, and <em>Z</em> = the symbol of the element .
Then your equation becomes
The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.
Then
55 = <em>x</em> - 1, so <em>x</em> = 55 + 1 = 56
135 = <em>y</em> + 0, so <em>y</em> = 135
Element 56 is barium, so the nuclear equation becomes
