Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states. Express your answer as a chemical equation. Identify all of the phases in your answer.
Write a balanced equation depicting the formation of one mole of SO3(g) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For SO3(g) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
For NaBr(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.
Write a balanced equation depicting the formation of one mole of Pb(NO3)2(s) from its elements in their standard states.
Express your answer as a chemical equation. Identify all of the phases in your answer.
For Pb(NO3)2(s) find the value of ΔH∘f. (Use Appendix C in the textbook.)
Express your answer using four significant figures.

Check Explanation.

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

NaBr (s)

The Standard formation reaction is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

Using appendix C, the standard heat of formation of NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The Standard formation reaction is

S (s) + (3/2) O₂ (g) → SO₃ (g)

Using appendix C, the standard heat of formation of SO₃(g) is

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The Standard formation reaction is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

Hope this Helps!!!

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally